题目：216. 组合总和 III

思路

回溯法

代码

class Solution {
private:vector<vector<int>> result;vector<int> path;public:void function(int k, int n, int startindex, int sum){int i;// 剪枝// 超过了, 不用找了;if(sum > n){return;}// 找到k个数了if(path.size() == k){if(sum == n){result.push_back(path);}return;}// 题目提示了从[1, 9]中找for(i = startindex; i <= 9; i++){// 剪枝// 需要找的数量 > 剩余的数量if((k - path.size()) > (9 - i + 1)){break;}sum += i;path.push_back(i);function(k, n, i+1, sum);sum -= i;path.pop_back();}}vector<vector<int>> combinationSum3(int k, int n) {result.clear();path.clear();function(k, n, 1, 0);return result;}
};

有2处剪枝哦！