题目来源：https://leetcode.cn/problems/combination-sum-ii/description/

 

C++题解： 这道题的难点在于解集中不能包含重复的组合。如果用set去重会造成超时，所以只能在单层递归逻辑中处理。通过识别下一个数与当前数是否相同，来修改下一次回溯的遍历起始值。

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int> candidates, int target, int startInd, int sum) {
        if(sum > target) return;
        else if(sum == target) {res.push_back(path); return;}

        for(int i = startInd; i < candidates.size(); i++) {
            if(sum + candidates[i] > target) break;
            path.push_back(candidates[i]);
            backtracking(candidates, target, i + 1, sum + candidates[i]);
            path.pop_back();
            while(i + 1 < candidates.size() && candidates[i] == candidates[i+1]) i++;
        }
        return;
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        if(candidates[0] > target) return res;
        backtracking(candidates, target, 0, 0);
        return res;
    }
};