[力扣题解]39. 组合总和

题目：39. 组合总和

思路

回溯法

代码，自己写的

// 自己写的
// 没有用 startindex 也可以实现！
class Solution {
public: vector<vector<int>> result;vector<int> path;void function(vector<int>& candidates, int target, int sum){int i;int size;if(sum == target){result.push_back(path);return;}// 因为没有负数, 所以不可能完成任务if(sum > target){return;}   size = candidates.size();for(i = 0; i < size; i++){// 后面找的数字不能比前面小if(!path.empty() && candidates[i] < path.back()){continue;}sum += candidates[i];path.push_back(candidates[i]);function(candidates, target, sum);sum -= candidates[i];path.pop_back();}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {sort(candidates.begin(), candidates.end());result.clear();path.clear();function(candidates, target, 0);return result;}
};

我没有用startindex，只是用if语句做了判断；
因为startindex的使用是为了避免重复，而本体允许重复，所以我认为可以不用，总的来说，我没有在for循环那一行上下功夫而是用if条件句来判断，保证不会出现这种情况：[2, 3, 2]，即后面找的数字比前面要小；

代码，来自代码随想录

class Solution {
private:vector<vector<int>> result;vector<int> path;void backtracking(vector<int>& candidates, int target, int sum, int startIndex) {if (sum == target) {result.push_back(path);return;}// 如果 sum + candidates[i] > target 就终止遍历for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {sum += candidates[i];path.push_back(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();sort(candidates.begin(), candidates.end()); // 需要排序backtracking(candidates, target, 0, 0);return result;}
};