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chapter_dynamic_programming/intro_to_dynamic_programming/index.html

@@ -3025,9 +3025,6 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
 <li>当 <span class="arithmatex">\(j\)</span> 等于 <span class="arithmatex">\(1\)</span> ，即上一轮跳了 <span class="arithmatex">\(1\)</span> 阶时，这一轮只能选择跳 <span class="arithmatex">\(2\)</span> 阶；</li>
 <li>当 <span class="arithmatex">\(j\)</span> 等于 <span class="arithmatex">\(2\)</span> ，即上一轮跳了 <span class="arithmatex">\(2\)</span> 阶时，这一轮可选择跳 <span class="arithmatex">\(1\)</span> 阶或跳 <span class="arithmatex">\(2\)</span> 阶；</li>
 </ul>
-<p><img alt="考虑约束下的递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_constraint_state_transfer.png" /></p>
-<p align="center"> Fig. 考虑约束下的递推关系 </p>
-
 <p>在该定义下，<span class="arithmatex">\(dp[i, j]\)</span> 表示状态 <span class="arithmatex">\([i, j]\)</span> 对应的方案数。由此，我们便能推导出以下的状态转移方程：</p>
 <div class="arithmatex">\[
 \begin{cases}
@@ -3035,6 +3032,9 @@ dp[i, 1] = dp[i-1, 2] \\
 dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
 \end{cases}
 \]</div>
+<p><img alt="考虑约束下的递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_constraint_state_transfer.png" /></p>
+<p align="center"> Fig. 考虑约束下的递推关系 </p>
+
 <p>最终，返回 <span class="arithmatex">\(dp[n, 1] + dp[n, 2]\)</span> 即可，两者之和代表爬到第 <span class="arithmatex">\(n\)</span> 阶的方案总数。</p>
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