给定一个数组 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用一次。
说明:
输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vis.resize(candidates.size());
        backTrace(candidates, target);
        return res;
    }
private:
    vector<vector<int>> res;
    vector<int> ans, vis;
    void backTrace(vector<int>& candidates, int target) {
        if (target == 0) {
            res.push_back(ans);
            return;
        }
        for (int i = 0; i < candidates.size(); i++) {
            if (vis[i]) continue;
            if (i > 0 && candidates[i] == candidates[i - 1] && vis[i - 1]) break;
            if (target < candidates[i]) break;
            vis[i] = 1;
            ans.push_back(candidates[i]);
            backTrace(candidates, target - candidates[i]);
            vis[i] = 0;
            ans.pop_back();
        }
    }
};
一会整理
原文:https://www.cnblogs.com/tmpUser/p/14502214.html