小山菌_代码随想录算法训练营第二十六天| 39. 组合总和、40.组合总和II 、 131.分割回文串

39. 组合总和

文档讲解：代码随想录.组合总和
视频讲解：带你学透回溯算法-组合总和（对应「leetcode」力扣题目：39.组合总和）| 回溯法精讲！
状态：已完成

代码实现

class Solution {private:vector<vector<int>> result;vector<int> path;void backtracking(vector<int> candidates, int target, int sum,int start_index) {if (sum > target) {return;}if (sum == target) {result.push_back(path);}for (int i = start_index; i < candidates.size(); i++) {path.push_back(candidates[i]);sum += candidates[i];backtracking(candidates, target, sum, i);sum -= candidates[i];path.pop_back();}}public:vector<vector<int>> combinationSum(vector<int>& candidates, int target) {result.clear();path.clear();backtracking(candidates, target, 0, 0);return result;}
};

心得体会

1.回溯思想的理解更加深入，逐渐上手

40.组合总和II

文档讲解：代码随想录.组合总和II
视频讲解：回溯算法中的去重，树层去重树枝去重，你弄清楚了没？| LeetCode:40.组合总和II
状态：已完成

代码实现

class Solution {
private:vector<vector<int>> result;vector<int> path;void backtracking(vector<int>& candidates, int target, int sum,int start_index, vector<bool>& used) {if (sum == target) {result.push_back(path);return;}for (int i = start_index;i < candidates.size() && sum + candidates[i] <= target; i++) {if (i > 0 && candidates[i] == candidates[i - 1] &&used[i - 1] == false) {continue;}sum += candidates[i];path.push_back(candidates[i]);used[i] = true;backtracking(candidates, target, sum, i+1, used);used[i] = false;sum -= candidates[i];path.pop_back();}}public:vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {vector<bool> used(candidates.size(), false);result.clear();path.clear();sort(candidates.begin(), candidates.end());backtracking(candidates, target, 0, 0, used);return result;}
};