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</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">subset_sum_i_naive.swift</span><pre><span></span><code><a id="__codelineno-8-1" name="__codelineno-8-1" href="#__codelineno-8-1"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="n">backtrack</span><span class="p">}</span>
<a id="__codelineno-8-2" name="__codelineno-8-2" href="#__codelineno-8-2"></a>
<a id="__codelineno-8-3" name="__codelineno-8-3" href="#__codelineno-8-3"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="n">subsetSumINaive</span><span class="p">}</span>
<div class="highlight"><span class="filename">subset_sum_i_naive.swift</span><pre><span></span><code><a id="__codelineno-8-1" name="__codelineno-8-1" href="#__codelineno-8-1"></a><span class="cm">/* 回溯算法：子集和 I */</span>
<a id="__codelineno-8-2" name="__codelineno-8-2" href="#__codelineno-8-2"></a><span class="kd">func</span> <span class="nf">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="kr">inout</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">target</span><span class="p">:</span> <span class="nb">Int</span><span class="p">,</span> <span class="n">total</span><span class="p">:</span> <span class="nb">Int</span><span class="p">,</span> <span class="n">choices</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">res</span><span class="p">:</span> <span class="kr">inout</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]])</span> <span class="p">{</span>
<a id="__codelineno-8-3" name="__codelineno-8-3" href="#__codelineno-8-3"></a> <span class="c1">// 子集和等于 target 时，记录解</span>
<a id="__codelineno-8-4" name="__codelineno-8-4" href="#__codelineno-8-4"></a> <span class="k">if</span> <span class="n">total</span> <span class="p">==</span> <span class="n">target</span> <span class="p">{</span>
<a id="__codelineno-8-5" name="__codelineno-8-5" href="#__codelineno-8-5"></a> <span class="n">res</span><span class="p">.</span><span class="n">append</span><span class="p">(</span><span class="n">state</span><span class="p">)</span>
<a id="__codelineno-8-6" name="__codelineno-8-6" href="#__codelineno-8-6"></a> <span class="k">return</span>
<a id="__codelineno-8-7" name="__codelineno-8-7" href="#__codelineno-8-7"></a> <span class="p">}</span>
<a id="__codelineno-8-8" name="__codelineno-8-8" href="#__codelineno-8-8"></a> <span class="c1">// 遍历所有选择</span>
<a id="__codelineno-8-9" name="__codelineno-8-9" href="#__codelineno-8-9"></a> <span class="k">for</span> <span class="n">i</span> <span class="k">in</span> <span class="bp">stride</span><span class="p">(</span><span class="n">from</span><span class="p">:</span> <span class="mi">0</span><span class="p">,</span> <span class="n">to</span><span class="p">:</span> <span class="n">choices</span><span class="p">.</span><span class="bp">count</span><span class="p">,</span> <span class="n">by</span><span class="p">:</span> <span class="mi">1</span><span class="p">)</span> <span class="p">{</span>
<a id="__codelineno-8-10" name="__codelineno-8-10" href="#__codelineno-8-10"></a> <span class="c1">// 剪枝：若子集和超过 target ，则跳过该选择</span>
<a id="__codelineno-8-11" name="__codelineno-8-11" href="#__codelineno-8-11"></a> <span class="k">if</span> <span class="n">total</span> <span class="o">+</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">&gt;</span> <span class="n">target</span> <span class="p">{</span>
<a id="__codelineno-8-12" name="__codelineno-8-12" href="#__codelineno-8-12"></a> <span class="k">continue</span>
<a id="__codelineno-8-13" name="__codelineno-8-13" href="#__codelineno-8-13"></a> <span class="p">}</span>
<a id="__codelineno-8-14" name="__codelineno-8-14" href="#__codelineno-8-14"></a> <span class="c1">// 尝试：做出选择，更新元素和 total</span>
<a id="__codelineno-8-15" name="__codelineno-8-15" href="#__codelineno-8-15"></a> <span class="n">state</span><span class="p">.</span><span class="n">append</span><span class="p">(</span><span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
<a id="__codelineno-8-16" name="__codelineno-8-16" href="#__codelineno-8-16"></a> <span class="c1">// 进行下一轮选择</span>
<a id="__codelineno-8-17" name="__codelineno-8-17" href="#__codelineno-8-17"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">state</span><span class="p">,</span> <span class="n">target</span><span class="p">:</span> <span class="n">target</span><span class="p">,</span> <span class="n">total</span><span class="p">:</span> <span class="n">total</span> <span class="o">+</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">choices</span><span class="p">:</span> <span class="n">choices</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">res</span><span class="p">)</span>
<a id="__codelineno-8-18" name="__codelineno-8-18" href="#__codelineno-8-18"></a> <span class="c1">// 回退：撤销选择，恢复到之前的状态</span>
<a id="__codelineno-8-19" name="__codelineno-8-19" href="#__codelineno-8-19"></a> <span class="n">state</span><span class="p">.</span><span class="bp">removeLast</span><span class="p">()</span>
<a id="__codelineno-8-20" name="__codelineno-8-20" href="#__codelineno-8-20"></a> <span class="p">}</span>
<a id="__codelineno-8-21" name="__codelineno-8-21" href="#__codelineno-8-21"></a><span class="p">}</span>
<a id="__codelineno-8-22" name="__codelineno-8-22" href="#__codelineno-8-22"></a>
<a id="__codelineno-8-23" name="__codelineno-8-23" href="#__codelineno-8-23"></a><span class="cm">/* 求解子集和 I（包含重复子集） */</span>
<a id="__codelineno-8-24" name="__codelineno-8-24" href="#__codelineno-8-24"></a><span class="kd">func</span> <span class="nf">subsetSumINaive</span><span class="p">(</span><span class="n">nums</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">target</span><span class="p">:</span> <span class="nb">Int</span><span class="p">)</span> <span class="p">-&gt;</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]]</span> <span class="p">{</span>
<a id="__codelineno-8-25" name="__codelineno-8-25" href="#__codelineno-8-25"></a> <span class="kd">var</span> <span class="nv">state</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">]</span> <span class="p">=</span> <span class="p">[]</span> <span class="c1">// 状态（子集）</span>
<a id="__codelineno-8-26" name="__codelineno-8-26" href="#__codelineno-8-26"></a> <span class="kd">let</span> <span class="nv">total</span> <span class="p">=</span> <span class="mi">0</span> <span class="c1">// 子集和</span>
<a id="__codelineno-8-27" name="__codelineno-8-27" href="#__codelineno-8-27"></a> <span class="kd">var</span> <span class="nv">res</span><span class="p">:</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]]</span> <span class="p">=</span> <span class="p">[]</span> <span class="c1">// 结果列表（子集列表）</span>
<a id="__codelineno-8-28" name="__codelineno-8-28" href="#__codelineno-8-28"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">state</span><span class="p">,</span> <span class="n">target</span><span class="p">:</span> <span class="n">target</span><span class="p">,</span> <span class="n">total</span><span class="p">:</span> <span class="n">total</span><span class="p">,</span> <span class="n">choices</span><span class="p">:</span> <span class="n">nums</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">res</span><span class="p">)</span>
<a id="__codelineno-8-29" name="__codelineno-8-29" href="#__codelineno-8-29"></a> <span class="k">return</span> <span class="n">res</span>
<a id="__codelineno-8-30" name="__codelineno-8-30" href="#__codelineno-8-30"></a><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
@ -2666,9 +2693,39 @@
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">subset_sum_i.swift</span><pre><span></span><code><a id="__codelineno-19-1" name="__codelineno-19-1" href="#__codelineno-19-1"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="n">backtrack</span><span class="p">}</span>
<a id="__codelineno-19-2" name="__codelineno-19-2" href="#__codelineno-19-2"></a>
<a id="__codelineno-19-3" name="__codelineno-19-3" href="#__codelineno-19-3"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="n">subsetSumI</span><span class="p">}</span>
<div class="highlight"><span class="filename">subset_sum_i.swift</span><pre><span></span><code><a id="__codelineno-19-1" name="__codelineno-19-1" href="#__codelineno-19-1"></a><span class="cm">/* 回溯算法：子集和 I */</span>
<a id="__codelineno-19-2" name="__codelineno-19-2" href="#__codelineno-19-2"></a><span class="kd">func</span> <span class="nf">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="kr">inout</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">target</span><span class="p">:</span> <span class="nb">Int</span><span class="p">,</span> <span class="n">choices</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">start</span><span class="p">:</span> <span class="nb">Int</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="kr">inout</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]])</span> <span class="p">{</span>
<a id="__codelineno-19-3" name="__codelineno-19-3" href="#__codelineno-19-3"></a> <span class="c1">// 子集和等于 target 时，记录解</span>
<a id="__codelineno-19-4" name="__codelineno-19-4" href="#__codelineno-19-4"></a> <span class="k">if</span> <span class="n">target</span> <span class="p">==</span> <span class="mi">0</span> <span class="p">{</span>
<a id="__codelineno-19-5" name="__codelineno-19-5" href="#__codelineno-19-5"></a> <span class="n">res</span><span class="p">.</span><span class="n">append</span><span class="p">(</span><span class="n">state</span><span class="p">)</span>
<a id="__codelineno-19-6" name="__codelineno-19-6" href="#__codelineno-19-6"></a> <span class="k">return</span>
<a id="__codelineno-19-7" name="__codelineno-19-7" href="#__codelineno-19-7"></a> <span class="p">}</span>
<a id="__codelineno-19-8" name="__codelineno-19-8" href="#__codelineno-19-8"></a> <span class="c1">// 遍历所有选择</span>
<a id="__codelineno-19-9" name="__codelineno-19-9" href="#__codelineno-19-9"></a> <span class="c1">// 剪枝二：从 start 开始遍历，避免生成重复子集</span>
<a id="__codelineno-19-10" name="__codelineno-19-10" href="#__codelineno-19-10"></a> <span class="k">for</span> <span class="n">i</span> <span class="k">in</span> <span class="bp">stride</span><span class="p">(</span><span class="n">from</span><span class="p">:</span> <span class="n">start</span><span class="p">,</span> <span class="n">to</span><span class="p">:</span> <span class="n">choices</span><span class="p">.</span><span class="bp">count</span><span class="p">,</span> <span class="n">by</span><span class="p">:</span> <span class="mi">1</span><span class="p">)</span> <span class="p">{</span>
<a id="__codelineno-19-11" name="__codelineno-19-11" href="#__codelineno-19-11"></a> <span class="c1">// 剪枝一：若子集和超过 target ，则直接结束循环</span>
<a id="__codelineno-19-12" name="__codelineno-19-12" href="#__codelineno-19-12"></a> <span class="c1">// 这是因为数组已排序，后边元素更大，子集和一定超过 target</span>
<a id="__codelineno-19-13" name="__codelineno-19-13" href="#__codelineno-19-13"></a> <span class="k">if</span> <span class="n">target</span> <span class="o">-</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">&lt;</span> <span class="mi">0</span> <span class="p">{</span>
<a id="__codelineno-19-14" name="__codelineno-19-14" href="#__codelineno-19-14"></a> <span class="k">break</span>
<a id="__codelineno-19-15" name="__codelineno-19-15" href="#__codelineno-19-15"></a> <span class="p">}</span>
<a id="__codelineno-19-16" name="__codelineno-19-16" href="#__codelineno-19-16"></a> <span class="c1">// 尝试：做出选择，更新 target, start</span>
<a id="__codelineno-19-17" name="__codelineno-19-17" href="#__codelineno-19-17"></a> <span class="n">state</span><span class="p">.</span><span class="n">append</span><span class="p">(</span><span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
<a id="__codelineno-19-18" name="__codelineno-19-18" href="#__codelineno-19-18"></a> <span class="c1">// 进行下一轮选择</span>
<a id="__codelineno-19-19" name="__codelineno-19-19" href="#__codelineno-19-19"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">state</span><span class="p">,</span> <span class="n">target</span><span class="p">:</span> <span class="n">target</span> <span class="o">-</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">choices</span><span class="p">:</span> <span class="n">choices</span><span class="p">,</span> <span class="n">start</span><span class="p">:</span> <span class="n">i</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">res</span><span class="p">)</span>
<a id="__codelineno-19-20" name="__codelineno-19-20" href="#__codelineno-19-20"></a> <span class="c1">// 回退：撤销选择，恢复到之前的状态</span>
<a id="__codelineno-19-21" name="__codelineno-19-21" href="#__codelineno-19-21"></a> <span class="n">state</span><span class="p">.</span><span class="bp">removeLast</span><span class="p">()</span>
<a id="__codelineno-19-22" name="__codelineno-19-22" href="#__codelineno-19-22"></a> <span class="p">}</span>
<a id="__codelineno-19-23" name="__codelineno-19-23" href="#__codelineno-19-23"></a><span class="p">}</span>
<a id="__codelineno-19-24" name="__codelineno-19-24" href="#__codelineno-19-24"></a>
<a id="__codelineno-19-25" name="__codelineno-19-25" href="#__codelineno-19-25"></a><span class="cm">/* 求解子集和 I */</span>
<a id="__codelineno-19-26" name="__codelineno-19-26" href="#__codelineno-19-26"></a><span class="kd">func</span> <span class="nf">subsetSumI</span><span class="p">(</span><span class="n">nums</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">target</span><span class="p">:</span> <span class="nb">Int</span><span class="p">)</span> <span class="p">-&gt;</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]]</span> <span class="p">{</span>
<a id="__codelineno-19-27" name="__codelineno-19-27" href="#__codelineno-19-27"></a> <span class="kd">var</span> <span class="nv">state</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">]</span> <span class="p">=</span> <span class="p">[]</span> <span class="c1">// 状态（子集）</span>
<a id="__codelineno-19-28" name="__codelineno-19-28" href="#__codelineno-19-28"></a> <span class="kd">let</span> <span class="nv">nums</span> <span class="p">=</span> <span class="n">nums</span><span class="p">.</span><span class="bp">sorted</span><span class="p">()</span> <span class="c1">// 对 nums 进行排序</span>
<a id="__codelineno-19-29" name="__codelineno-19-29" href="#__codelineno-19-29"></a> <span class="kd">let</span> <span class="nv">start</span> <span class="p">=</span> <span class="mi">0</span> <span class="c1">// 遍历起始点</span>
<a id="__codelineno-19-30" name="__codelineno-19-30" href="#__codelineno-19-30"></a> <span class="kd">var</span> <span class="nv">res</span><span class="p">:</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]]</span> <span class="p">=</span> <span class="p">[]</span> <span class="c1">// 结果列表（子集列表）</span>
<a id="__codelineno-19-31" name="__codelineno-19-31" href="#__codelineno-19-31"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">state</span><span class="p">,</span> <span class="n">target</span><span class="p">:</span> <span class="n">target</span><span class="p">,</span> <span class="n">choices</span><span class="p">:</span> <span class="n">nums</span><span class="p">,</span> <span class="n">start</span><span class="p">:</span> <span class="n">start</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">res</span><span class="p">)</span>
<a id="__codelineno-19-32" name="__codelineno-19-32" href="#__codelineno-19-32"></a> <span class="k">return</span> <span class="n">res</span>
<a id="__codelineno-19-33" name="__codelineno-19-33" href="#__codelineno-19-33"></a><span class="p">}</span>
</code></pre></div>
</div>
<div class="tabbed-block">
@ -2923,9 +2980,44 @@
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><span class="filename">subset_sum_ii.swift</span><pre><span></span><code><a id="__codelineno-30-1" name="__codelineno-30-1" href="#__codelineno-30-1"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="n">backtrack</span><span class="p">}</span>
<a id="__codelineno-30-2" name="__codelineno-30-2" href="#__codelineno-30-2"></a>
<a id="__codelineno-30-3" name="__codelineno-30-3" href="#__codelineno-30-3"></a><span class="p">[</span><span class="kd">class</span><span class="p">]{}</span><span class="o">-</span><span class="p">[</span><span class="kd">func</span><span class="p">]{</span><span class="n">subsetSumII</span><span class="p">}</span>
<div class="highlight"><span class="filename">subset_sum_ii.swift</span><pre><span></span><code><a id="__codelineno-30-1" name="__codelineno-30-1" href="#__codelineno-30-1"></a><span class="cm">/* 回溯算法：子集和 II */</span>
<a id="__codelineno-30-2" name="__codelineno-30-2" href="#__codelineno-30-2"></a><span class="kd">func</span> <span class="nf">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="kr">inout</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">target</span><span class="p">:</span> <span class="nb">Int</span><span class="p">,</span> <span class="n">choices</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">start</span><span class="p">:</span> <span class="nb">Int</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="kr">inout</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]])</span> <span class="p">{</span>
<a id="__codelineno-30-3" name="__codelineno-30-3" href="#__codelineno-30-3"></a> <span class="c1">// 子集和等于 target 时，记录解</span>
<a id="__codelineno-30-4" name="__codelineno-30-4" href="#__codelineno-30-4"></a> <span class="k">if</span> <span class="n">target</span> <span class="p">==</span> <span class="mi">0</span> <span class="p">{</span>
<a id="__codelineno-30-5" name="__codelineno-30-5" href="#__codelineno-30-5"></a> <span class="n">res</span><span class="p">.</span><span class="n">append</span><span class="p">(</span><span class="n">state</span><span class="p">)</span>
<a id="__codelineno-30-6" name="__codelineno-30-6" href="#__codelineno-30-6"></a> <span class="k">return</span>
<a id="__codelineno-30-7" name="__codelineno-30-7" href="#__codelineno-30-7"></a> <span class="p">}</span>
<a id="__codelineno-30-8" name="__codelineno-30-8" href="#__codelineno-30-8"></a> <span class="c1">// 遍历所有选择</span>
<a id="__codelineno-30-9" name="__codelineno-30-9" href="#__codelineno-30-9"></a> <span class="c1">// 剪枝二：从 start 开始遍历，避免生成重复子集</span>
<a id="__codelineno-30-10" name="__codelineno-30-10" href="#__codelineno-30-10"></a> <span class="c1">// 剪枝三：从 start 开始遍历，避免重复选择同一元素</span>
<a id="__codelineno-30-11" name="__codelineno-30-11" href="#__codelineno-30-11"></a> <span class="k">for</span> <span class="n">i</span> <span class="k">in</span> <span class="bp">stride</span><span class="p">(</span><span class="n">from</span><span class="p">:</span> <span class="n">start</span><span class="p">,</span> <span class="n">to</span><span class="p">:</span> <span class="n">choices</span><span class="p">.</span><span class="bp">count</span><span class="p">,</span> <span class="n">by</span><span class="p">:</span> <span class="mi">1</span><span class="p">)</span> <span class="p">{</span>
<a id="__codelineno-30-12" name="__codelineno-30-12" href="#__codelineno-30-12"></a> <span class="c1">// 剪枝一：若子集和超过 target ，则直接结束循环</span>
<a id="__codelineno-30-13" name="__codelineno-30-13" href="#__codelineno-30-13"></a> <span class="c1">// 这是因为数组已排序，后边元素更大，子集和一定超过 target</span>
<a id="__codelineno-30-14" name="__codelineno-30-14" href="#__codelineno-30-14"></a> <span class="k">if</span> <span class="n">target</span> <span class="o">-</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">&lt;</span> <span class="mi">0</span> <span class="p">{</span>
<a id="__codelineno-30-15" name="__codelineno-30-15" href="#__codelineno-30-15"></a> <span class="k">break</span>
<a id="__codelineno-30-16" name="__codelineno-30-16" href="#__codelineno-30-16"></a> <span class="p">}</span>
<a id="__codelineno-30-17" name="__codelineno-30-17" href="#__codelineno-30-17"></a> <span class="c1">// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过</span>
<a id="__codelineno-30-18" name="__codelineno-30-18" href="#__codelineno-30-18"></a> <span class="k">if</span> <span class="n">i</span> <span class="o">&gt;</span> <span class="n">start</span><span class="p">,</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="p">==</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]</span> <span class="p">{</span>
<a id="__codelineno-30-19" name="__codelineno-30-19" href="#__codelineno-30-19"></a> <span class="k">continue</span>
<a id="__codelineno-30-20" name="__codelineno-30-20" href="#__codelineno-30-20"></a> <span class="p">}</span>
<a id="__codelineno-30-21" name="__codelineno-30-21" href="#__codelineno-30-21"></a> <span class="c1">// 尝试：做出选择，更新 target, start</span>
<a id="__codelineno-30-22" name="__codelineno-30-22" href="#__codelineno-30-22"></a> <span class="n">state</span><span class="p">.</span><span class="n">append</span><span class="p">(</span><span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
<a id="__codelineno-30-23" name="__codelineno-30-23" href="#__codelineno-30-23"></a> <span class="c1">// 进行下一轮选择</span>
<a id="__codelineno-30-24" name="__codelineno-30-24" href="#__codelineno-30-24"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">state</span><span class="p">,</span> <span class="n">target</span><span class="p">:</span> <span class="n">target</span> <span class="o">-</span> <span class="n">choices</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">choices</span><span class="p">:</span> <span class="n">choices</span><span class="p">,</span> <span class="n">start</span><span class="p">:</span> <span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">res</span><span class="p">)</span>
<a id="__codelineno-30-25" name="__codelineno-30-25" href="#__codelineno-30-25"></a> <span class="c1">// 回退：撤销选择，恢复到之前的状态</span>
<a id="__codelineno-30-26" name="__codelineno-30-26" href="#__codelineno-30-26"></a> <span class="n">state</span><span class="p">.</span><span class="bp">removeLast</span><span class="p">()</span>
<a id="__codelineno-30-27" name="__codelineno-30-27" href="#__codelineno-30-27"></a> <span class="p">}</span>
<a id="__codelineno-30-28" name="__codelineno-30-28" href="#__codelineno-30-28"></a><span class="p">}</span>
<a id="__codelineno-30-29" name="__codelineno-30-29" href="#__codelineno-30-29"></a>
<a id="__codelineno-30-30" name="__codelineno-30-30" href="#__codelineno-30-30"></a><span class="cm">/* 求解子集和 II */</span>
<a id="__codelineno-30-31" name="__codelineno-30-31" href="#__codelineno-30-31"></a><span class="kd">func</span> <span class="nf">subsetSumII</span><span class="p">(</span><span class="n">nums</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">],</span> <span class="n">target</span><span class="p">:</span> <span class="nb">Int</span><span class="p">)</span> <span class="p">-&gt;</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]]</span> <span class="p">{</span>
<a id="__codelineno-30-32" name="__codelineno-30-32" href="#__codelineno-30-32"></a> <span class="kd">var</span> <span class="nv">state</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">]</span> <span class="p">=</span> <span class="p">[]</span> <span class="c1">// 状态（子集）</span>
<a id="__codelineno-30-33" name="__codelineno-30-33" href="#__codelineno-30-33"></a> <span class="kd">let</span> <span class="nv">nums</span> <span class="p">=</span> <span class="n">nums</span><span class="p">.</span><span class="bp">sorted</span><span class="p">()</span> <span class="c1">// 对 nums 进行排序</span>
<a id="__codelineno-30-34" name="__codelineno-30-34" href="#__codelineno-30-34"></a> <span class="kd">let</span> <span class="nv">start</span> <span class="p">=</span> <span class="mi">0</span> <span class="c1">// 遍历起始点</span>
<a id="__codelineno-30-35" name="__codelineno-30-35" href="#__codelineno-30-35"></a> <span class="kd">var</span> <span class="nv">res</span><span class="p">:</span> <span class="p">[[</span><span class="nb">Int</span><span class="p">]]</span> <span class="p">=</span> <span class="p">[]</span> <span class="c1">// 结果列表（子集列表）</span>
<a id="__codelineno-30-36" name="__codelineno-30-36" href="#__codelineno-30-36"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">state</span><span class="p">,</span> <span class="n">target</span><span class="p">:</span> <span class="n">target</span><span class="p">,</span> <span class="n">choices</span><span class="p">:</span> <span class="n">nums</span><span class="p">,</span> <span class="n">start</span><span class="p">:</span> <span class="n">start</span><span class="p">,</span> <span class="n">res</span><span class="p">:</span> <span class="p">&amp;</span><span class="n">res</span><span class="p">)</span>
<a id="__codelineno-30-37" name="__codelineno-30-37" href="#__codelineno-30-37"></a> <span class="k">return</span> <span class="n">res</span>
<a id="__codelineno-30-38" name="__codelineno-30-38" href="#__codelineno-30-38"></a><span class="p">}</span>
</code></pre></div>
</div>
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<li class="md-nav__item">
<a href="../../chapter_dynamic_programming/dp_solution_pipeline.md" class="md-nav__link">
13.3. &nbsp; DP 解题步骤（New）
<a href="../../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
13.3. &nbsp; DP 解题思路（New）
</a>
</li>

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13.3. &nbsp; DP 解题步骤（New）
<a href="../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
13.3. &nbsp; DP 解题思路（New）
</a>
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13.3. &nbsp; DP 解题步骤（New）
<a href="../../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
13.3. &nbsp; DP 解题思路（New）
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13.3. &nbsp; DP 解题步骤（New）
<a href="../../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
13.3. &nbsp; DP 解题思路（New）
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13.3. &nbsp; DP 解题步骤（New）
<a href="../../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
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13.3. &nbsp; DP 解题步骤（New）
<a href="../../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
13.3. &nbsp; DP 解题思路（New）
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<a href="../../chapter_dynamic_programming/dp_solution_pipeline.md" class="md-nav__link">
13.3. &nbsp; DP 解题步骤（New）
<a href="../../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
13.3. &nbsp; DP 解题思路（New）
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13.3. &nbsp; DP 解题步骤（New）
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13.3. &nbsp; DP 解题步骤（New）
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13.3. &nbsp; DP 解题思路（New）
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<li class="md-nav__item">
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13.3. &nbsp; DP 解题步骤（New）
<a href="../chapter_dynamic_programming/dp_solution_pipeline/" class="md-nav__link">
13.3. &nbsp; DP 解题思路（New）
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13.3. &nbsp; DP 解题步骤（New）
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13.3. &nbsp; DP 解题思路（New）
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@ -2010,8 +2010,8 @@
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@ -2244,7 +2244,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">cost</span><span class="p">.</span><span class="na">length</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span>
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">cost</span><span class="o">[</span><span class="n">n</span><span class="o">]</span><span class="p">;</span>
<a id="__codelineno-0-6" name="__codelineno-0-6" href="#__codelineno-0-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-0-6" name="__codelineno-0-6" href="#__codelineno-0-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-0-7" name="__codelineno-0-7" href="#__codelineno-0-7"></a><span class="w"> </span><span class="kt">int</span><span class="o">[]</span><span class="w"> </span><span class="n">dp</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="o">[</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="mi">1</span><span class="o">]</span><span class="p">;</span>
<a id="__codelineno-0-8" name="__codelineno-0-8" href="#__codelineno-0-8"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-0-9" name="__codelineno-0-9" href="#__codelineno-0-9"></a><span class="w"> </span><span class="n">dp</span><span class="o">[</span><span class="mi">1</span><span class="o">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">cost</span><span class="o">[</span><span class="mi">1</span><span class="o">]</span><span class="p">;</span>
@ -2263,7 +2263,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">cost</span><span class="p">.</span><span class="n">size</span><span class="p">()</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span>
<a id="__codelineno-1-5" name="__codelineno-1-5" href="#__codelineno-1-5"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">cost</span><span class="p">[</span><span class="n">n</span><span class="p">];</span>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-1-7" name="__codelineno-1-7" href="#__codelineno-1-7"></a><span class="w"> </span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">dp</span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="mi">1</span><span class="p">);</span>
<a id="__codelineno-1-8" name="__codelineno-1-8" href="#__codelineno-1-8"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-1-9" name="__codelineno-1-9" href="#__codelineno-1-9"></a><span class="w"> </span><span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">cost</span><span class="p">[</span><span class="mi">1</span><span class="p">];</span>
@ -2282,7 +2282,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
<a id="__codelineno-2-3" name="__codelineno-2-3" href="#__codelineno-2-3"></a> <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">cost</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
<a id="__codelineno-2-4" name="__codelineno-2-4" href="#__codelineno-2-4"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">1</span> <span class="ow">or</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">2</span><span class="p">:</span>
<a id="__codelineno-2-5" name="__codelineno-2-5" href="#__codelineno-2-5"></a> <span class="k">return</span> <span class="n">cost</span><span class="p">[</span><span class="n">n</span><span class="p">]</span>
<a id="__codelineno-2-6" name="__codelineno-2-6" href="#__codelineno-2-6"></a> <span class="c1"># 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-2-6" name="__codelineno-2-6" href="#__codelineno-2-6"></a> <span class="c1"># 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-2-7" name="__codelineno-2-7" href="#__codelineno-2-7"></a> <span class="n">dp</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
<a id="__codelineno-2-8" name="__codelineno-2-8" href="#__codelineno-2-8"></a> <span class="c1"># 初始状态：预设最小子问题的解</span>
<a id="__codelineno-2-9" name="__codelineno-2-9" href="#__codelineno-2-9"></a> <span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">],</span> <span class="n">dp</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span> <span class="o">=</span> <span class="n">cost</span><span class="p">[</span><span class="mi">1</span><span class="p">],</span> <span class="n">cost</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span>
@ -2314,7 +2314,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
<a id="__codelineno-7-3" name="__codelineno-7-3" href="#__codelineno-7-3"></a><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">cost</span><span class="p">.</span><span class="n">Length</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="m">1</span><span class="p">;</span>
<a id="__codelineno-7-4" name="__codelineno-7-4" href="#__codelineno-7-4"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="m">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="m">2</span><span class="p">)</span>
<a id="__codelineno-7-5" name="__codelineno-7-5" href="#__codelineno-7-5"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">cost</span><span class="p">[</span><span class="n">n</span><span class="p">];</span>
<a id="__codelineno-7-6" name="__codelineno-7-6" href="#__codelineno-7-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-7-6" name="__codelineno-7-6" href="#__codelineno-7-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-7-7" name="__codelineno-7-7" href="#__codelineno-7-7"></a><span class="w"> </span><span class="kt">int</span><span class="p">[]</span><span class="w"> </span><span class="n">dp</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="p">[</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="m">1</span><span class="p">];</span>
<a id="__codelineno-7-8" name="__codelineno-7-8" href="#__codelineno-7-8"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-7-9" name="__codelineno-7-9" href="#__codelineno-7-9"></a><span class="w"> </span><span class="n">dp</span><span class="p">[</span><span class="m">1</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">cost</span><span class="p">[</span><span class="m">1</span><span class="p">];</span>
@ -2475,7 +2475,7 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
<a id="__codelineno-22-3" name="__codelineno-22-3" href="#__codelineno-22-3"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-22-4" name="__codelineno-22-4" href="#__codelineno-22-4"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">n</span><span class="p">;</span>
<a id="__codelineno-22-5" name="__codelineno-22-5" href="#__codelineno-22-5"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-22-6" name="__codelineno-22-6" href="#__codelineno-22-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-22-6" name="__codelineno-22-6" href="#__codelineno-22-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-22-7" name="__codelineno-22-7" href="#__codelineno-22-7"></a><span class="w"> </span><span class="kt">int</span><span class="o">[][]</span><span class="w"> </span><span class="n">dp</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="o">[</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="mi">1</span><span class="o">][</span><span class="mi">3</span><span class="o">]</span><span class="p">;</span>
<a id="__codelineno-22-8" name="__codelineno-22-8" href="#__codelineno-22-8"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-22-9" name="__codelineno-22-9" href="#__codelineno-22-9"></a><span class="w"> </span><span class="n">dp</span><span class="o">[</span><span class="mi">1</span><span class="o">][</span><span class="mi">1</span><span class="o">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
@ -2497,7 +2497,7 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
<a id="__codelineno-23-3" name="__codelineno-23-3" href="#__codelineno-23-3"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-23-4" name="__codelineno-23-4" href="#__codelineno-23-4"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">n</span><span class="p">;</span>
<a id="__codelineno-23-5" name="__codelineno-23-5" href="#__codelineno-23-5"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-23-6" name="__codelineno-23-6" href="#__codelineno-23-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-23-6" name="__codelineno-23-6" href="#__codelineno-23-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-23-7" name="__codelineno-23-7" href="#__codelineno-23-7"></a><span class="w"> </span><span class="n">vector</span><span class="o">&lt;</span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;&gt;</span><span class="w"> </span><span class="n">dp</span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span><span class="w"> </span><span class="mi">0</span><span class="p">));</span>
<a id="__codelineno-23-8" name="__codelineno-23-8" href="#__codelineno-23-8"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-23-9" name="__codelineno-23-9" href="#__codelineno-23-9"></a><span class="w"> </span><span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">][</span><span class="mi">1</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
@ -2518,7 +2518,7 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
<a id="__codelineno-24-2" name="__codelineno-24-2" href="#__codelineno-24-2"></a><span class="w"> </span><span class="sd">&quot;&quot;&quot;带约束爬楼梯：动态规划&quot;&quot;&quot;</span>
<a id="__codelineno-24-3" name="__codelineno-24-3" href="#__codelineno-24-3"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">1</span> <span class="ow">or</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">2</span><span class="p">:</span>
<a id="__codelineno-24-4" name="__codelineno-24-4" href="#__codelineno-24-4"></a> <span class="k">return</span> <span class="n">n</span>
<a id="__codelineno-24-5" name="__codelineno-24-5" href="#__codelineno-24-5"></a> <span class="c1"># 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-24-5" name="__codelineno-24-5" href="#__codelineno-24-5"></a> <span class="c1"># 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-24-6" name="__codelineno-24-6" href="#__codelineno-24-6"></a> <span class="n">dp</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="mi">3</span> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)]</span>
<a id="__codelineno-24-7" name="__codelineno-24-7" href="#__codelineno-24-7"></a> <span class="c1"># 初始状态：预设最小子问题的解</span>
<a id="__codelineno-24-8" name="__codelineno-24-8" href="#__codelineno-24-8"></a> <span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">][</span><span class="mi">1</span><span class="p">],</span> <span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">][</span><span class="mi">2</span><span class="p">]</span> <span class="o">=</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">0</span>
@ -2552,7 +2552,7 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
<a id="__codelineno-29-3" name="__codelineno-29-3" href="#__codelineno-29-3"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="m">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="m">2</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-29-4" name="__codelineno-29-4" href="#__codelineno-29-4"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">n</span><span class="p">;</span>
<a id="__codelineno-29-5" name="__codelineno-29-5" href="#__codelineno-29-5"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-29-6" name="__codelineno-29-6" href="#__codelineno-29-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-29-6" name="__codelineno-29-6" href="#__codelineno-29-6"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-29-7" name="__codelineno-29-7" href="#__codelineno-29-7"></a><span class="w"> </span><span class="kt">int</span><span class="p">[,]</span><span class="w"> </span><span class="n">dp</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="p">[</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="m">1</span><span class="p">,</span><span class="w"> </span><span class="m">3</span><span class="p">];</span>
<a id="__codelineno-29-8" name="__codelineno-29-8" href="#__codelineno-29-8"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-29-9" name="__codelineno-29-9" href="#__codelineno-29-9"></a><span class="w"> </span><span class="n">dp</span><span class="p">[</span><span class="m">1</span><span class="p">,</span><span class="w"> </span><span class="m">1</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">1</span><span class="p">;</span>
@ -2682,13 +2682,13 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
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@ -2689,7 +2689,7 @@ dp[i] = dp[i-1] + dp[i-2]
<a id="__codelineno-33-2" name="__codelineno-33-2" href="#__codelineno-33-2"></a><span class="kt">int</span><span class="w"> </span><span class="nf">climbingStairsDP</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-33-3" name="__codelineno-33-3" href="#__codelineno-33-3"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span>
<a id="__codelineno-33-4" name="__codelineno-33-4" href="#__codelineno-33-4"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">n</span><span class="p">;</span>
<a id="__codelineno-33-5" name="__codelineno-33-5" href="#__codelineno-33-5"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-33-5" name="__codelineno-33-5" href="#__codelineno-33-5"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-33-6" name="__codelineno-33-6" href="#__codelineno-33-6"></a><span class="w"> </span><span class="kt">int</span><span class="o">[]</span><span class="w"> </span><span class="n">dp</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="o">[</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="mi">1</span><span class="o">]</span><span class="p">;</span>
<a id="__codelineno-33-7" name="__codelineno-33-7" href="#__codelineno-33-7"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-33-8" name="__codelineno-33-8" href="#__codelineno-33-8"></a><span class="w"> </span><span class="n">dp</span><span class="o">[</span><span class="mi">1</span><span class="o">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
@ -2707,7 +2707,7 @@ dp[i] = dp[i-1] + dp[i-2]
<a id="__codelineno-34-2" name="__codelineno-34-2" href="#__codelineno-34-2"></a><span class="kt">int</span><span class="w"> </span><span class="nf">climbingStairsDP</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-34-3" name="__codelineno-34-3" href="#__codelineno-34-3"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="mi">2</span><span class="p">)</span>
<a id="__codelineno-34-4" name="__codelineno-34-4" href="#__codelineno-34-4"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">n</span><span class="p">;</span>
<a id="__codelineno-34-5" name="__codelineno-34-5" href="#__codelineno-34-5"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-34-5" name="__codelineno-34-5" href="#__codelineno-34-5"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-34-6" name="__codelineno-34-6" href="#__codelineno-34-6"></a><span class="w"> </span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">dp</span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="mi">1</span><span class="p">);</span>
<a id="__codelineno-34-7" name="__codelineno-34-7" href="#__codelineno-34-7"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-34-8" name="__codelineno-34-8" href="#__codelineno-34-8"></a><span class="w"> </span><span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">1</span><span class="p">;</span>
@ -2725,7 +2725,7 @@ dp[i] = dp[i-1] + dp[i-2]
<a id="__codelineno-35-2" name="__codelineno-35-2" href="#__codelineno-35-2"></a><span class="w"> </span><span class="sd">&quot;&quot;&quot;爬楼梯：动态规划&quot;&quot;&quot;</span>
<a id="__codelineno-35-3" name="__codelineno-35-3" href="#__codelineno-35-3"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">1</span> <span class="ow">or</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">2</span><span class="p">:</span>
<a id="__codelineno-35-4" name="__codelineno-35-4" href="#__codelineno-35-4"></a> <span class="k">return</span> <span class="n">n</span>
<a id="__codelineno-35-5" name="__codelineno-35-5" href="#__codelineno-35-5"></a> <span class="c1"># 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-35-5" name="__codelineno-35-5" href="#__codelineno-35-5"></a> <span class="c1"># 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-35-6" name="__codelineno-35-6" href="#__codelineno-35-6"></a> <span class="n">dp</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
<a id="__codelineno-35-7" name="__codelineno-35-7" href="#__codelineno-35-7"></a> <span class="c1"># 初始状态：预设最小子问题的解</span>
<a id="__codelineno-35-8" name="__codelineno-35-8" href="#__codelineno-35-8"></a> <span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">],</span> <span class="n">dp</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span> <span class="o">=</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">2</span>
@ -2756,7 +2756,7 @@ dp[i] = dp[i-1] + dp[i-2]
<a id="__codelineno-40-2" name="__codelineno-40-2" href="#__codelineno-40-2"></a><span class="kt">int</span><span class="w"> </span><span class="nf">climbingStairsDP</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-40-3" name="__codelineno-40-3" href="#__codelineno-40-3"></a><span class="w"> </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="m">1</span><span class="w"> </span><span class="o">||</span><span class="w"> </span><span class="n">n</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="m">2</span><span class="p">)</span>
<a id="__codelineno-40-4" name="__codelineno-40-4" href="#__codelineno-40-4"></a><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">n</span><span class="p">;</span>
<a id="__codelineno-40-5" name="__codelineno-40-5" href="#__codelineno-40-5"></a><span class="w"> </span><span class="c1">// 初始化 dp 列表，用于存储子问题的解</span>
<a id="__codelineno-40-5" name="__codelineno-40-5" href="#__codelineno-40-5"></a><span class="w"> </span><span class="c1">// 初始化 dp 表，用于存储子问题的解</span>
<a id="__codelineno-40-6" name="__codelineno-40-6" href="#__codelineno-40-6"></a><span class="w"> </span><span class="kt">int</span><span class="p">[]</span><span class="w"> </span><span class="n">dp</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="p">[</span><span class="n">n</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="m">1</span><span class="p">];</span>
<a id="__codelineno-40-7" name="__codelineno-40-7" href="#__codelineno-40-7"></a><span class="w"> </span><span class="c1">// 初始状态：预设最小子问题的解</span>
<a id="__codelineno-40-8" name="__codelineno-40-8" href="#__codelineno-40-8"></a><span class="w"> </span><span class="n">dp</span><span class="p">[</span><span class="m">1</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">1</span><span class="p">;</span>
@ -2785,14 +2785,14 @@ dp[i] = dp[i-1] + dp[i-2]
</div>
<p>与回溯算法一样，动态规划也使用“状态”概念来表示问题求解的某个特定阶段，每个状态都对应一个子问题以及相应的局部最优解。例如对于爬楼梯问题，状态定义为当前所在楼梯阶数 <span class="arithmatex">\(i\)</span> 。<strong>动态规划的常用术语包括</strong>：</p>
<ul>
<li>将 <span class="arithmatex">\(dp\)</span> 数组称为「状态列表」，<span class="arithmatex">\(dp[i]\)</span> 代表第 <span class="arithmatex">\(i\)</span> 个状态的解；</li>
<li>将数组 <code>dp</code> 称为「<span class="arithmatex">\(dp\)</span> 表」，<span class="arithmatex">\(dp[i]\)</span> 表示状态 <span class="arithmatex">\(i\)</span> 对应子问题的解；</li>
<li>将最小子问题对应的状态（即第 <span class="arithmatex">\(1\)</span> , <span class="arithmatex">\(2\)</span> 阶楼梯）称为「初始状态」；</li>
<li>将递推公式 <span class="arithmatex">\(dp[i] = dp[i-1] + dp[i-2]\)</span> 称为「状态转移方程」；</li>
</ul>
<p><img alt="爬楼梯的动态规划过程" src="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" /></p>
<p align="center"> Fig. 爬楼梯的动态规划过程 </p>
<p>细心的你可能发现，<strong>由于 <span class="arithmatex">\(dp[i]\)</span> 只与 <span class="arithmatex">\(dp[i-1]\)</span> 和 <span class="arithmatex">\(dp[i-2]\)</span> 有关，因此我们无需使用一个数组 <code>dp</code> 来存储所有状态</strong>，而只需两个变量滚动前进即可。如以下代码所示，由于省去了数组 <code>dp</code> 占用的空间，因此空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
<p>细心的你可能发现，<strong>由于 <span class="arithmatex">\(dp[i]\)</span> 只与 <span class="arithmatex">\(dp[i-1]\)</span> 和 <span class="arithmatex">\(dp[i-2]\)</span> 有关，因此我们无需使用一个数组 <code>dp</code> 来存储所有子问题的解</strong>，而只需两个变量滚动前进即可。如以下代码所示，由于省去了数组 <code>dp</code> 占用的空间，因此空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
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@ -15,7 +15,7 @@
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@ -1962,8 +1962,8 @@
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@ -2242,12 +2242,11 @@
<p align="center"> Fig. 0-1 背包的示例数据 </p>
<p>我们可以将 0-1 背包问题看作是一个由 <span class="arithmatex">\(n\)</span> 轮决策组成的过程，每个物体都有不放入和放入两种决策，因此该问题是满足决策树模型的。此外，该问题的目标是求解“在限定背包容量下的最大价值”，因此较大概率是个动态规划问题。我们接下来尝试求解它。</p>
<p><strong>第一步：思考每一轮的决策是什么，从而得到状态定义</strong></p>
<p>在 0-1 背包问题中，不放入背包，背包容量不变；放入背包，背包容量减小。由此可得状态定义：物品编号 <span class="arithmatex">\(i\)</span> 和背包容量 <span class="arithmatex">\(c\)</span> ，记为 <span class="arithmatex">\([i, c]\)</span> 。</p>
<p><strong>第二步：明确子问题是什么，从而得到 <span class="arithmatex">\(dp\)</span> 列表</strong></p>
<p>状态 <span class="arithmatex">\([i, c]\)</span> 对应的子问题为：<strong>前 <span class="arithmatex">\(i\)</span> 个物品在容量为 <span class="arithmatex">\(c\)</span> 背包中的最大价值</strong>，记为 <span class="arithmatex">\(dp[i, c]\)</span> 。</p>
<p><strong>第一步：思考每轮的决策，定义状态，从而得到 <span class="arithmatex">\(dp\)</span> 表</strong></p>
<p>在 0-1 背包问题中，不放入背包，背包容量不变；放入背包，背包容量减小。由此可得状态定义：当前物品编号 <span class="arithmatex">\(i\)</span> 和剩余背包容量 <span class="arithmatex">\(c\)</span> ，记为 <span class="arithmatex">\([i, c]\)</span> 。</p>
<p>状态 <span class="arithmatex">\([i, c]\)</span> 对应的子问题为：<strong>前 <span class="arithmatex">\(i\)</span> 个物品在剩余容量为 <span class="arithmatex">\(c\)</span> 的背包中的最大价值</strong>，记为 <span class="arithmatex">\(dp[i, c]\)</span> 。</p>
<p>至此，我们得到一个尺寸为 <span class="arithmatex">\(n \times cap\)</span> 的二维 <span class="arithmatex">\(dp\)</span> 矩阵。</p>
<p><strong>第三步：找出最优子结构，进而推导出状态转移方程</strong></p>
<p><strong>第二步：找出最优子结构，进而推导出状态转移方程</strong></p>
<p>当我们做出物品 <span class="arithmatex">\(i\)</span> 的决策后，剩余的是前 <span class="arithmatex">\(i-1\)</span> 个物品的决策。因此，状态转移分为两种情况：</p>
<ul>
<li><strong>不放入物品 <span class="arithmatex">\(i\)</span></strong> ：背包容量不变，状态转移至 <span class="arithmatex">\([i-1, c]\)</span> ；</li>
@ -2258,6 +2257,9 @@
dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
\]</div>
<p>需要注意的是，若当前物品重量 <span class="arithmatex">\(wgt[i - 1]\)</span> 超出剩余背包容量 <span class="arithmatex">\(c\)</span> ，则只能选择不放入背包。</p>
<p><strong>第三步：确定边界条件和状态转移顺序</strong></p>
<p>当无物品或无剩余背包容量时最大价值为 <span class="arithmatex">\(0\)</span> ，即所有 <span class="arithmatex">\(dp[i, 0]\)</span> 和 <span class="arithmatex">\(dp[0, c]\)</span> 都等于 <span class="arithmatex">\(0\)</span> 。</p>
<p>当前状态 <span class="arithmatex">\([i, c]\)</span> 从上方的状态 <span class="arithmatex">\([i-1, c]\)</span> 和左上方的状态 <span class="arithmatex">\([i-1, c-wgt[i-1]]\)</span> 转移而来，因此通过两层循环正序遍历整个 <span class="arithmatex">\(dp\)</span> 表即可。</p>
<h2 id="1341">13.4.1. &nbsp; 方法一：暴力搜索<a class="headerlink" href="#1341" title="Permanent link">&para;</a></h2>
<p>搜索代码包含以下要素：</p>
<ul>
@ -2416,7 +2418,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<div class="highlight"><span class="filename">knapsack.py</span><pre><span></span><code><a id="__codelineno-24-1" name="__codelineno-24-1" href="#__codelineno-24-1"></a><span class="k">def</span> <span class="nf">knapsack_dp</span><span class="p">(</span><span class="n">wgt</span><span class="p">,</span> <span class="n">val</span><span class="p">,</span> <span class="n">cap</span><span class="p">):</span>
<a id="__codelineno-24-2" name="__codelineno-24-2" href="#__codelineno-24-2"></a><span class="w"> </span><span class="sd">&quot;&quot;&quot;0-1 背包：动态规划&quot;&quot;&quot;</span>
<a id="__codelineno-24-3" name="__codelineno-24-3" href="#__codelineno-24-3"></a> <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">wgt</span><span class="p">)</span>
<a id="__codelineno-24-4" name="__codelineno-24-4" href="#__codelineno-24-4"></a> <span class="c1"># 初始化 dp 列表</span>
<a id="__codelineno-24-4" name="__codelineno-24-4" href="#__codelineno-24-4"></a> <span class="c1"># 初始化 dp 表</span>
<a id="__codelineno-24-5" name="__codelineno-24-5" href="#__codelineno-24-5"></a> <span class="n">dp</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">cap</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)]</span>
<a id="__codelineno-24-6" name="__codelineno-24-6" href="#__codelineno-24-6"></a> <span class="c1"># 状态转移</span>
<a id="__codelineno-24-7" name="__codelineno-24-7" href="#__codelineno-24-7"></a> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">):</span>
@ -2551,7 +2553,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<div class="highlight"><span class="filename">knapsack.py</span><pre><span></span><code><a id="__codelineno-35-1" name="__codelineno-35-1" href="#__codelineno-35-1"></a><span class="k">def</span> <span class="nf">knapsack_dp_comp</span><span class="p">(</span><span class="n">wgt</span><span class="p">,</span> <span class="n">val</span><span class="p">,</span> <span class="n">cap</span><span class="p">):</span>
<a id="__codelineno-35-2" name="__codelineno-35-2" href="#__codelineno-35-2"></a><span class="w"> </span><span class="sd">&quot;&quot;&quot;0-1 背包：状态压缩后的动态规划&quot;&quot;&quot;</span>
<a id="__codelineno-35-3" name="__codelineno-35-3" href="#__codelineno-35-3"></a> <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">wgt</span><span class="p">)</span>
<a id="__codelineno-35-4" name="__codelineno-35-4" href="#__codelineno-35-4"></a> <span class="c1"># 初始化 dp 列表</span>
<a id="__codelineno-35-4" name="__codelineno-35-4" href="#__codelineno-35-4"></a> <span class="c1"># 初始化 dp 表</span>
<a id="__codelineno-35-5" name="__codelineno-35-5" href="#__codelineno-35-5"></a> <span class="n">dp</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">cap</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
<a id="__codelineno-35-6" name="__codelineno-35-6" href="#__codelineno-35-6"></a> <span class="c1"># 状态转移</span>
<a id="__codelineno-35-7" name="__codelineno-35-7" href="#__codelineno-35-7"></a> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">):</span>
@ -2677,7 +2679,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<nav class="md-footer__inner md-grid" aria-label="页脚" >
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@ -2686,7 +2688,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
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