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Add the section of subset sum problem. (#558)
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16 changed files with 821 additions and 5 deletions
57
codes/cpp/chapter_backtracking/subset_sum_i.cpp
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57
codes/cpp/chapter_backtracking/subset_sum_i.cpp
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/**
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* File: subset_sum_i.cpp
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* Created Time: 2023-06-21
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 回溯算法:子集和 I */
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void backtrack(vector<int> &state, int target, vector<int> &choices, int start, vector<vector<int>> &res) {
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// 子集和等于 target 时,记录解
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if (target == 0) {
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res.push_back(state);
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return;
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}
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// 遍历所有选择
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// 剪枝二:从 start 开始遍历,避免生成重复子集
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for (int i = start; i < choices.size(); i++) {
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// 剪枝一:若子集和超过 target ,则直接结束循环
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// 这是因为数组已排序,后边元素更大,子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 尝试:做出选择,更新 target, start
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state.push_back(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i, res);
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// 回退:撤销选择,恢复到之前的状态
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state.pop_back();
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}
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}
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/* 求解子集和 I */
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vector<vector<int>> subsetSumI(vector<int> &nums, int target) {
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vector<int> state; // 状态(子集)
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sort(nums.begin(), nums.end()); // 对 nums 进行排序
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int start = 0; // 遍历起始点
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vector<vector<int>> res; // 结果列表(子集列表)
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backtrack(state, target, nums, start, res);
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return res;
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}
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/* Driver Code */
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int main() {
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vector<int> nums = {3, 4, 5};
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int target = 9;
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vector<vector<int>> res = subsetSumI(nums, target);
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cout << "输入数组 nums = ";
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printVector(nums);
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cout << "target = " << target << endl;
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cout << "所有和等于 " << target << " 的子集 res = " << endl;
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printVectorMatrix(res);
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return 0;
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}
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54
codes/cpp/chapter_backtracking/subset_sum_i_naive.cpp
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54
codes/cpp/chapter_backtracking/subset_sum_i_naive.cpp
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@ -0,0 +1,54 @@
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/**
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* File: subset_sum_i_naive.cpp
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* Created Time: 2023-06-21
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 回溯算法:子集和 I */
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void backtrack(vector<int> &state, int target, int total, vector<int> &choices, vector<vector<int>> &res) {
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// 子集和等于 target 时,记录解
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if (total == target) {
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res.push_back(state);
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return;
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}
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// 遍历所有选择
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for (size_t i = 0; i < choices.size(); i++) {
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// 剪枝:若子集和超过 target ,则跳过该选择
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if (total + choices[i] > target) {
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continue;
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}
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// 尝试:做出选择,更新元素和 total
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state.push_back(choices[i]);
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// 进行下一轮选择
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backtrack(state, target, total + choices[i], choices, res);
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// 回退:撤销选择,恢复到之前的状态
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state.pop_back();
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}
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}
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/* 求解子集和 I(包含重复子集) */
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vector<vector<int>> subsetSumINaive(vector<int> &nums, int target) {
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vector<int> state; // 状态(子集)
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int total = 0; // 子集和
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vector<vector<int>> res; // 结果列表(子集列表)
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backtrack(state, target, total, nums, res);
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return res;
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}
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/* Driver Code */
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int main() {
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vector<int> nums = {3, 4, 5};
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int target = 9;
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vector<vector<int>> res = subsetSumINaive(nums, target);
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cout << "输入数组 nums = ";
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printVector(nums);
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cout << "target = " << target << endl;
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cout << "所有和等于 " << target << " 的子集 res = " << endl;
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printVectorMatrix(res);
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return 0;
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}
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62
codes/cpp/chapter_backtracking/subset_sum_ii.cpp
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62
codes/cpp/chapter_backtracking/subset_sum_ii.cpp
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/**
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* File: subset_sum_ii.cpp
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* Created Time: 2023-06-21
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 回溯算法:子集和 II */
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void backtrack(vector<int> &state, int target, vector<int> &choices, int start, vector<vector<int>> &res) {
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// 子集和等于 target 时,记录解
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if (target == 0) {
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res.push_back(state);
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return;
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}
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// 遍历所有选择
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// 剪枝二:从 start 开始遍历,避免生成重复子集
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// 剪枝三:从 start 开始遍历,避免重复选择同一元素
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for (int i = start; i < choices.size(); i++) {
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// 剪枝一:若子集和超过 target ,则直接结束循环
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// 这是因为数组已排序,后边元素更大,子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 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
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if (i > start && choices[i] == choices[i - 1]) {
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continue;
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}
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// 尝试:做出选择,更新 target, start
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state.push_back(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i + 1, res);
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// 回退:撤销选择,恢复到之前的状态
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state.pop_back();
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}
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}
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/* 求解子集和 II */
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vector<vector<int>> subsetSumII(vector<int> &nums, int target) {
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vector<int> state; // 状态(子集)
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sort(nums.begin(), nums.end()); // 对 nums 进行排序
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int start = 0; // 遍历起始点
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vector<vector<int>> res; // 结果列表(子集列表)
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backtrack(state, target, nums, start, res);
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return res;
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}
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/* Driver Code */
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int main() {
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vector<int> nums = {4, 4, 5};
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int target = 9;
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vector<vector<int>> res = subsetSumII(nums, target);
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cout << "输入数组 nums = ";
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printVector(nums);
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cout << "target = " << target << endl;
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cout << "所有和等于 " << target << " 的子集 res = " << endl;
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printVectorMatrix(res);
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return 0;
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}
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55
codes/java/chapter_backtracking/subset_sum_i.java
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55
codes/java/chapter_backtracking/subset_sum_i.java
Normal file
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/**
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* File: subset_sum_i.java
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* Created Time: 2023-06-21
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* Author: Krahets (krahets@163.com)
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*/
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package chapter_backtracking;
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import java.util.*;
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public class subset_sum_i {
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/* 回溯算法:子集和 I */
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static void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
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// 子集和等于 target 时,记录解
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if (target == 0) {
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res.add(new ArrayList<>(state));
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return;
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}
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// 遍历所有选择
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// 剪枝二:从 start 开始遍历,避免生成重复子集
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for (int i = start; i < choices.length; i++) {
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// 剪枝一:若子集和超过 target ,则直接结束循环
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// 这是因为数组已排序,后边元素更大,子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 尝试:做出选择,更新 target, start
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state.add(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i, res);
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// 回退:撤销选择,恢复到之前的状态
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state.remove(state.size() - 1);
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}
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}
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/* 求解子集和 I */
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static List<List<Integer>> subsetSumI(int[] nums, int target) {
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List<Integer> state = new ArrayList<>(); // 状态(子集)
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Arrays.sort(nums); // 对 nums 进行排序
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int start = 0; // 遍历起始点
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List<List<Integer>> res = new ArrayList<>(); // 结果列表(子集列表)
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backtrack(state, target, nums, start, res);
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return res;
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}
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public static void main(String[] args) {
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int[] nums = { 3, 4, 5 };
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int target = 9;
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List<List<Integer>> res = subsetSumI(nums, target);
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System.out.println("输入数组 nums = " + Arrays.toString(nums) + ", target = " + target);
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System.out.println("所有和等于 " + target + " 的子集 res = " + res);
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}
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}
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53
codes/java/chapter_backtracking/subset_sum_i_naive.java
Normal file
53
codes/java/chapter_backtracking/subset_sum_i_naive.java
Normal file
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@ -0,0 +1,53 @@
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/**
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* File: subset_sum_i_naive.java
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* Created Time: 2023-06-21
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* Author: Krahets (krahets@163.com)
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*/
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package chapter_backtracking;
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import java.util.*;
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public class subset_sum_i_naive {
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/* 回溯算法:子集和 I */
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static void backtrack(List<Integer> state, int target, int total, int[] choices, List<List<Integer>> res) {
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// 子集和等于 target 时,记录解
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if (total == target) {
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res.add(new ArrayList<>(state));
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return;
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}
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// 遍历所有选择
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for (int i = 0; i < choices.length; i++) {
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// 剪枝:若子集和超过 target ,则跳过该选择
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if (total + choices[i] > target) {
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continue;
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}
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// 尝试:做出选择,更新元素和 total
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state.add(choices[i]);
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// 进行下一轮选择
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backtrack(state, target, total + choices[i], choices, res);
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// 回退:撤销选择,恢复到之前的状态
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state.remove(state.size() - 1);
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}
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}
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/* 求解子集和 I(包含重复子集) */
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static List<List<Integer>> subsetSumINaive(int[] nums, int target) {
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List<Integer> state = new ArrayList<>(); // 状态(子集)
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int total = 0; // 子集和
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List<List<Integer>> res = new ArrayList<>(); // 结果列表(子集列表)
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backtrack(state, target, total, nums, res);
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return res;
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}
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public static void main(String[] args) {
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int[] nums = { 3, 4, 5 };
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int target = 9;
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List<List<Integer>> res = subsetSumINaive(nums, target);
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System.out.println("输入数组 nums = " + Arrays.toString(nums) + ", target = " + target);
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System.out.println("所有和等于 " + target + " 的子集 res = " + res);
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System.out.println("请注意,该方法输出的结果包含重复集合");
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}
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}
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60
codes/java/chapter_backtracking/subset_sum_ii.java
Normal file
60
codes/java/chapter_backtracking/subset_sum_ii.java
Normal file
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@ -0,0 +1,60 @@
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/**
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* File: subset_sum_ii.java
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* Created Time: 2023-06-21
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* Author: Krahets (krahets@163.com)
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*/
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package chapter_backtracking;
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import java.util.*;
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public class subset_sum_ii {
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/* 回溯算法:子集和 II */
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static void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
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// 子集和等于 target 时,记录解
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if (target == 0) {
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res.add(new ArrayList<>(state));
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return;
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}
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// 遍历所有选择
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// 剪枝二:从 start 开始遍历,避免生成重复子集
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// 剪枝三:从 start 开始遍历,避免重复选择同一元素
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for (int i = start; i < choices.length; i++) {
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// 剪枝一:若子集和超过 target ,则直接结束循环
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// 这是因为数组已排序,后边元素更大,子集和一定超过 target
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if (target - choices[i] < 0) {
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break;
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}
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// 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
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if (i > start && choices[i] == choices[i - 1]) {
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continue;
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}
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// 尝试:做出选择,更新 target, start
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state.add(choices[i]);
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// 进行下一轮选择
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backtrack(state, target - choices[i], choices, i + 1, res);
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// 回退:撤销选择,恢复到之前的状态
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state.remove(state.size() - 1);
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}
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}
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/* 求解子集和 II */
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static List<List<Integer>> subsetSumII(int[] nums, int target) {
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List<Integer> state = new ArrayList<>(); // 状态(子集)
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Arrays.sort(nums); // 对 nums 进行排序
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int start = 0; // 遍历起始点
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List<List<Integer>> res = new ArrayList<>(); // 结果列表(子集列表)
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backtrack(state, target, nums, start, res);
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return res;
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}
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public static void main(String[] args) {
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int[] nums = { 4, 4, 5 };
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int target = 9;
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List<List<Integer>> res = subsetSumII(nums, target);
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System.out.println("输入数组 nums = " + Arrays.toString(nums) + ", target = " + target);
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System.out.println("所有和等于 " + target + " 的子集 res = " + res);
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}
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}
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48
codes/python/chapter_backtracking/subset_sum_i.py
Normal file
48
codes/python/chapter_backtracking/subset_sum_i.py
Normal file
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"""
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File: subset_sum_i.py
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Created Time: 2023-06-17
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Author: Krahets (krahets@163.com)
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"""
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def backtrack(
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state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
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):
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"""回溯算法:子集和 I"""
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# 子集和等于 target 时,记录解
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if target == 0:
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res.append(list(state))
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return
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# 遍历所有选择
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# 剪枝二:从 start 开始遍历,避免生成重复子集
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for i in range(start, len(choices)):
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# 剪枝一:若子集和超过 target ,则直接结束循环
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# 这是因为数组已排序,后边元素更大,子集和一定超过 target
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if target - choices[i] < 0:
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break
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# 尝试:做出选择,更新 target, start
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state.append(choices[i])
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# 进行下一轮选择
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backtrack(state, target - choices[i], choices, i, res)
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# 回退:撤销选择,恢复到之前的状态
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state.pop()
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def subset_sum_i(nums: list[int], target: int) -> list[list[int]]:
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"""求解子集和 I"""
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state = [] # 状态(子集)
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nums.sort() # 对 nums 进行排序
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start = 0 # 遍历起始点
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res = [] # 结果列表(子集列表)
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backtrack(state, target, nums, start, res)
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return res
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"""Driver Code"""
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if __name__ == "__main__":
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nums = [3, 4, 5]
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target = 9
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res = subset_sum_i(nums, target)
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print(f"输入数组 nums = {nums}, target = {target}")
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print(f"所有和等于 {target} 的子集 res = {res}")
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50
codes/python/chapter_backtracking/subset_sum_i_naive.py
Normal file
50
codes/python/chapter_backtracking/subset_sum_i_naive.py
Normal file
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"""
|
||||
File: subset_sum_i_naive.py
|
||||
Created Time: 2023-06-17
|
||||
Author: Krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
|
||||
def backtrack(
|
||||
state: list[int],
|
||||
target: int,
|
||||
total: int,
|
||||
choices: list[int],
|
||||
res: list[list[int]],
|
||||
):
|
||||
"""回溯算法:子集和 I"""
|
||||
# 子集和等于 target 时,记录解
|
||||
if total == target:
|
||||
res.append(list(state))
|
||||
return
|
||||
# 遍历所有选择
|
||||
for i in range(len(choices)):
|
||||
# 剪枝:若子集和超过 target ,则跳过该选择
|
||||
if total + choices[i] > target:
|
||||
continue
|
||||
# 尝试:做出选择,更新元素和 total
|
||||
state.append(choices[i])
|
||||
# 进行下一轮选择
|
||||
backtrack(state, target, total + choices[i], choices, res)
|
||||
# 回退:撤销选择,恢复到之前的状态
|
||||
state.pop()
|
||||
|
||||
|
||||
def subset_sum_i_naive(nums: list[int], target: int) -> list[list[int]]:
|
||||
"""求解子集和 I(包含重复子集)"""
|
||||
state = [] # 状态(子集)
|
||||
total = 0 # 子集和
|
||||
res = [] # 结果列表(子集列表)
|
||||
backtrack(state, target, total, nums, res)
|
||||
return res
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
nums = [3, 4, 5]
|
||||
target = 9
|
||||
res = subset_sum_i_naive(nums, target)
|
||||
|
||||
print(f"输入数组 nums = {nums}, target = {target}")
|
||||
print(f"所有和等于 {target} 的子集 res = {res}")
|
||||
print(f"请注意,该方法输出的结果包含重复集合")
|
52
codes/python/chapter_backtracking/subset_sum_ii.py
Normal file
52
codes/python/chapter_backtracking/subset_sum_ii.py
Normal file
|
@ -0,0 +1,52 @@
|
|||
"""
|
||||
File: subset_sum_ii.py
|
||||
Created Time: 2023-06-17
|
||||
Author: Krahets (krahets@163.com)
|
||||
"""
|
||||
|
||||
|
||||
def backtrack(
|
||||
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
|
||||
):
|
||||
"""回溯算法:子集和 II"""
|
||||
# 子集和等于 target 时,记录解
|
||||
if target == 0:
|
||||
res.append(list(state))
|
||||
return
|
||||
# 遍历所有选择
|
||||
# 剪枝二:从 start 开始遍历,避免生成重复子集
|
||||
# 剪枝三:从 start 开始遍历,避免重复选择同一元素
|
||||
for i in range(start, len(choices)):
|
||||
# 剪枝一:若子集和超过 target ,则直接结束循环
|
||||
# 这是因为数组已排序,后边元素更大,子集和一定超过 target
|
||||
if target - choices[i] < 0:
|
||||
break
|
||||
# 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
|
||||
if i > start and choices[i] == choices[i - 1]:
|
||||
continue
|
||||
# 尝试:做出选择,更新 target, start
|
||||
state.append(choices[i])
|
||||
# 进行下一轮选择
|
||||
backtrack(state, target - choices[i], choices, i + 1, res)
|
||||
# 回退:撤销选择,恢复到之前的状态
|
||||
state.pop()
|
||||
|
||||
|
||||
def subset_sum_ii(nums: list[int], target: int) -> list[list[int]]:
|
||||
"""求解子集和 II"""
|
||||
state = [] # 状态(子集)
|
||||
nums.sort() # 对 nums 进行排序
|
||||
start = 0 # 遍历起始点
|
||||
res = [] # 结果列表(子集列表)
|
||||
backtrack(state, target, nums, start, res)
|
||||
return res
|
||||
|
||||
|
||||
"""Driver Code"""
|
||||
if __name__ == "__main__":
|
||||
nums = [4, 4, 5]
|
||||
target = 9
|
||||
res = subset_sum_ii(nums, target)
|
||||
|
||||
print(f"输入数组 nums = {nums}, target = {target}")
|
||||
print(f"所有和等于 {target} 的子集 res = {res}")
|
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324
docs/chapter_backtracking/subset_sum_problem.md
Normal file
324
docs/chapter_backtracking/subset_sum_problem.md
Normal file
|
@ -0,0 +1,324 @@
|
|||
# 子集和问题
|
||||
|
||||
!!! question
|
||||
|
||||
给定一个正整数数组 `nums` 和一个目标正整数 `target` ,请找出所有可能的组合,使得组合中的元素和等于 `target` 。给定数组无重复元素,每个元素可以被选取多次。请以列表形式返回这些组合,列表中不应包含重复组合。
|
||||
|
||||
例如,输入集合 $\{3, 4, 5\}$ 和目标整数 $9$ ,由于集合中的数字可以被重复选取,因此解为 $\{3, 3, 3\}, \{4, 5\}$ 。请注意,子集是不区分元素顺序的,例如 $\{4, 5\}$ 和 $\{5, 4\}$ 是同一个子集。
|
||||
|
||||
## 参考全排列解法
|
||||
|
||||
从回溯算法的角度看,我们可以把子集的生成过程想象成一系列选择的结果,并在选择过程中实时更新“元素和”,当元素和等于 `target` 时,就将子集记录至结果列表。
|
||||
|
||||
与上节全排列问题不同的是,本题允许重复选取同一元素,因此无需借助 `selected` 布尔列表来记录元素是否已被选择。我们可以对全排列代码进行小幅修改,初步得到解题代码。
|
||||
|
||||
=== "Java"
|
||||
|
||||
```java title="subset_sum_i_naive.java"
|
||||
[class]{subset_sum_i_naive}-[func]{backtrack}
|
||||
|
||||
[class]{subset_sum_i_naive}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "C++"
|
||||
|
||||
```cpp title="subset_sum_i_naive.cpp"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "Python"
|
||||
|
||||
```python title="subset_sum_i_naive.py"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subset_sum_i_naive}
|
||||
```
|
||||
|
||||
=== "Go"
|
||||
|
||||
```go title="subset_sum_i_naive.go"
|
||||
[class]{}-[func]{backtrackINaive}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "JavaScript"
|
||||
|
||||
```javascript title="subset_sum_i_naive.js"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "TypeScript"
|
||||
|
||||
```typescript title="subset_sum_i_naive.ts"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "C"
|
||||
|
||||
```c title="subset_sum_i_naive.c"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "C#"
|
||||
|
||||
```csharp title="subset_sum_i_naive.cs"
|
||||
[class]{subset_sum_i_naive}-[func]{backtrack}
|
||||
|
||||
[class]{subset_sum_i_naive}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "Swift"
|
||||
|
||||
```swift title="subset_sum_i_naive.swift"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "Zig"
|
||||
|
||||
```zig title="subset_sum_i_naive.zig"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
=== "Dart"
|
||||
|
||||
```dart title="subset_sum_i_naive.dart"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumINaive}
|
||||
```
|
||||
|
||||
向以上代码输入数组 $[3, 4, 5]$ 和目标元素 $9$ ,输出结果为 $[3, 3, 3], [4, 5], [5, 4]$ 。**虽然成功找出了所有和为 $9$ 的子集,但其中存在重复的子集 $[4, 5]$ 和 $[5, 4]$** 。这是因为搜索过程是区分选择顺序的,如下图所示,先选 $4$ 后选 $5$ 与先选 $5$ 后选 $4$ 是两种不同的情况。
|
||||
|
||||
![子集搜索与越界剪枝](subset_sum_problem.assets/subset_sum_i_naive.png)
|
||||
|
||||
## 重复子集剪枝
|
||||
|
||||
为了去除重复子集,**一种直接的思路是对结果列表进行去重**。但这个方法效率很低,因为:
|
||||
|
||||
- 当数组元素较多,尤其是当 `target` 较大时,搜索过程会产生大量的重复子集。
|
||||
- 比较子集(数组)的异同是很耗时的,需要先排序数组,再比较数组中每个元素的异同。
|
||||
|
||||
为了达到最佳效率,**我们希望在搜索过程中通过剪枝进行去重**。观察下图,重复子集是在以不同顺序选择数组元素时产生的,具体来看:
|
||||
|
||||
1. 第一轮和第二轮分别选择 $3$ , $4$ ,会生成包含这两个元素的所有子集,记为 $[3, 4, \cdots]$ 。
|
||||
2. 若第一轮选择 $4$ ,**则第二轮应该跳过 $3$** ,因为该选择产生的子集 $[4, 3, \cdots]$ 和 `1.` 中提到的子集完全重复。
|
||||
3. 同理,若第一轮选择 $5$ ,**则第二轮应该跳过 $3$ 和 $4$** ,因为子集 $[5, 3, \cdots]$ 和子集 $[5, 4, \cdots]$ 和之前的子集重复。
|
||||
|
||||
![不同选择顺序导致的重复子集](subset_sum_problem.assets/subset_sum_i_pruning.png)
|
||||
|
||||
总结来看,给定输入数组 $[x_1, x_2, \cdots, x_n]$ ,设搜索过程中的选择序列为 $[x_{i_1}, x_{i_2}, \cdots , x_{i_m}]$ ,则该选择序列需要满足 $i_1 \leq i_2 \leq \cdots \leq i_m$ 。**不满足该条件的选择序列都是重复子集**。
|
||||
|
||||
为实现该剪枝,我们初始化变量 `start` ,用于指示遍历起点。**当做出选择 $x_{i}$ 后,设定下一轮从索引 $i$ 开始遍历**,从而完成子集去重。
|
||||
|
||||
除此之外,我们还对代码进行了两项优化。首先,我们在开启搜索前将数组 `nums` 排序,在搜索过程中,**当子集和超过 `target` 时直接结束循环**,因为后边的元素更大,其子集和都一定会超过 `target` 。其次,**我们通过在 `target` 上执行减法来统计元素和**,当 `target` 等于 $0$ 时记录解,省去了元素和变量 `total` 。
|
||||
|
||||
=== "Java"
|
||||
|
||||
```java title="subset_sum_i.java"
|
||||
[class]{subset_sum_i}-[func]{backtrack}
|
||||
|
||||
[class]{subset_sum_i}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "C++"
|
||||
|
||||
```cpp title="subset_sum_i.cpp"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "Python"
|
||||
|
||||
```python title="subset_sum_i.py"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subset_sum_i}
|
||||
```
|
||||
|
||||
=== "Go"
|
||||
|
||||
```go title="subset_sum_i.go"
|
||||
[class]{}-[func]{backtrackI}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "JavaScript"
|
||||
|
||||
```javascript title="subset_sum_i.js"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "TypeScript"
|
||||
|
||||
```typescript title="subset_sum_i.ts"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "C"
|
||||
|
||||
```c title="subset_sum_i.c"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "C#"
|
||||
|
||||
```csharp title="subset_sum_i.cs"
|
||||
[class]{subset_sum_i}-[func]{backtrack}
|
||||
|
||||
[class]{subset_sum_i}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "Swift"
|
||||
|
||||
```swift title="subset_sum_i.swift"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "Zig"
|
||||
|
||||
```zig title="subset_sum_i.zig"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
=== "Dart"
|
||||
|
||||
```dart title="subset_sum_i.dart"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumI}
|
||||
```
|
||||
|
||||
如下图所示,为将数组 $[3, 4, 5]$ 和目标元素 $9$ 输入到以上代码后的整体回溯过程。
|
||||
|
||||
![子集和 I 回溯过程](subset_sum_problem.assets/subset_sum_i.png)
|
||||
|
||||
## 考虑相等元素
|
||||
|
||||
!!! question
|
||||
|
||||
给定一个正整数数组 `nums` 和一个目标正整数 `target` ,请找出所有可能的组合,使得组合中的元素和等于 `target` 。**给定数组可能包含重复元素,每个元素只可被选择一次**。请以列表形式返回这些组合,列表中不应包含重复组合。
|
||||
|
||||
相比于上题,**本题的输入数组可能包含重复元素**,这引入了新的问题。例如,给定数组 $[4, \hat{4}, 5]$ 和目标元素 $9$ ,则现有代码的输出结果为 $[4, 5], [\hat{4}, 5]$ ,也出现了重复子集。**造成这种重复的原因是相等元素在某轮中被多次选择**。如下图所示,第一轮共有三个选择,其中两个都为 $4$ ,会产生两个重复的搜索分支,从而输出重复子集;同理,第二轮的两个 $4$ 也会产生重复子集。
|
||||
|
||||
![相等元素导致的重复子集](subset_sum_problem.assets/subset_sum_ii_repeat.png)
|
||||
|
||||
为解决此问题,**我们需要限制相等元素在每一轮中只被选择一次**。实现方式比较巧妙:由于数组是已排序的,因此相等元素都是相邻的。利用该特性,在某轮选择中,若当前元素与其左边元素相等,则说明它已经被选择过,因此直接跳过当前元素。
|
||||
|
||||
与此同时,**本题规定数组元素只能被选择一次**。幸运的是,我们也可以利用变量 `start` 来满足该约束:当做出选择 $x_{i}$ 后,设定下一轮从索引 $i + 1$ 开始向后遍历。这样即能去除重复子集,也能避免重复选择相等元素。
|
||||
|
||||
=== "Java"
|
||||
|
||||
```java title="subset_sum_ii.java"
|
||||
[class]{subset_sum_ii}-[func]{backtrack}
|
||||
|
||||
[class]{subset_sum_ii}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "C++"
|
||||
|
||||
```cpp title="subset_sum_ii.cpp"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "Python"
|
||||
|
||||
```python title="subset_sum_ii.py"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subset_sum_ii}
|
||||
```
|
||||
|
||||
=== "Go"
|
||||
|
||||
```go title="subset_sum_ii.go"
|
||||
[class]{}-[func]{backtrackII}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "JavaScript"
|
||||
|
||||
```javascript title="subset_sum_ii.js"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "TypeScript"
|
||||
|
||||
```typescript title="subset_sum_ii.ts"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "C"
|
||||
|
||||
```c title="subset_sum_ii.c"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "C#"
|
||||
|
||||
```csharp title="subset_sum_ii.cs"
|
||||
[class]{subset_sum_ii}-[func]{backtrack}
|
||||
|
||||
[class]{subset_sum_ii}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "Swift"
|
||||
|
||||
```swift title="subset_sum_ii.swift"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "Zig"
|
||||
|
||||
```zig title="subset_sum_ii.zig"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
=== "Dart"
|
||||
|
||||
```dart title="subset_sum_ii.dart"
|
||||
[class]{}-[func]{backtrack}
|
||||
|
||||
[class]{}-[func]{subsetSumII}
|
||||
```
|
||||
|
||||
下图展示了数组 $[4, 4, 5]$ 和目标元素 $9$ 的回溯过程,共包含四种剪枝操作。建议你将图示与代码注释相结合,理解整个搜索过程,以及每种剪枝操作是如何工作的。
|
||||
|
||||
![子集和 II 回溯过程](subset_sum_problem.assets/subset_sum_ii.png)
|
11
mkdocs.yml
11
mkdocs.yml
|
@ -187,20 +187,20 @@ nav:
|
|||
- 9.4. 小结: chapter_graph/summary.md
|
||||
- 10. 搜索:
|
||||
- chapter_searching/index.md
|
||||
- 10.1. 二分查找(New): chapter_searching/binary_search.md
|
||||
- 10.2. 二分查找边界(New): chapter_searching/binary_search_edge.md
|
||||
- 10.1. 二分查找: chapter_searching/binary_search.md
|
||||
- 10.2. 二分查找边界: chapter_searching/binary_search_edge.md
|
||||
- 10.3. 哈希优化策略: chapter_searching/replace_linear_by_hashing.md
|
||||
- 10.4. 重识搜索算法: chapter_searching/searching_algorithm_revisited.md
|
||||
- 10.5. 小结: chapter_searching/summary.md
|
||||
- 11. 排序:
|
||||
- chapter_sorting/index.md
|
||||
- 11.1. 排序算法: chapter_sorting/sorting_algorithm.md
|
||||
- 11.2. 选择排序(New): chapter_sorting/selection_sort.md
|
||||
- 11.2. 选择排序: chapter_sorting/selection_sort.md
|
||||
- 11.3. 冒泡排序: chapter_sorting/bubble_sort.md
|
||||
- 11.4. 插入排序: chapter_sorting/insertion_sort.md
|
||||
- 11.5. 快速排序: chapter_sorting/quick_sort.md
|
||||
- 11.6. 归并排序: chapter_sorting/merge_sort.md
|
||||
- 11.7. 堆排序(New): chapter_sorting/heap_sort.md
|
||||
- 11.7. 堆排序: chapter_sorting/heap_sort.md
|
||||
- 11.8. 桶排序: chapter_sorting/bucket_sort.md
|
||||
- 11.9. 计数排序: chapter_sorting/counting_sort.md
|
||||
- 11.10. 基数排序: chapter_sorting/radix_sort.md
|
||||
|
@ -209,7 +209,8 @@ nav:
|
|||
- chapter_backtracking/index.md
|
||||
- 12.1. 回溯算法: chapter_backtracking/backtracking_algorithm.md
|
||||
- 12.2. 全排列问题: chapter_backtracking/permutations_problem.md
|
||||
- 12.3. N 皇后问题: chapter_backtracking/n_queens_problem.md
|
||||
- 12.3. 子集和问题(New): chapter_backtracking/subset_sum_problem.md
|
||||
- 12.4. N 皇后问题: chapter_backtracking/n_queens_problem.md
|
||||
- 13. 附录:
|
||||
- 13.1. 编程环境安装: chapter_appendix/installation.md
|
||||
- 13.2. 一起参与创作: chapter_appendix/contribution.md
|
||||
|
|
Loading…
Reference in a new issue