leetCode 组合数II 40

这道题和以往不同的是要去重

//关键
              if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false) {
                continue;
              }
//如果used上一个used为true表示包含在其中，可以取,如果false，肯会重复，如下图

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