力扣算法题—040组合求和二

#include "000库函数.h"

//第一感觉使用回溯比较快
//96ms


class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>>R;
        sort(candidates.begin(), candidates.end());
        if (candidates.size() == 0)return R;
        vector<int>v;//临时存放解    
        Combin(candidates, R, v, target, 0, 0);        
        return R;
    }

    void Combin(vector<int>candidates, vector<vector<int>>&Res, vector<int>&v, int target, int start, int sum) {
        if (sum == target) {
            //sort(v.begin(), v.end());
            Res.push_back(v);
            return;
        }
        else if (sum > target)
            return;

        for (int i = start; i < candidates.size(); ++i) {
            if (i > start&&candidates[i] == candidates[i - 1])continue;//去除重复的组合            
            v.push_back(candidates[i]);
            Combin(candidates, Res, v, target, i + 1, sum + candidates[i]);//i+1是与上题的不同之处，不会出现重复使用元素
            v.pop_back();//将数字退出                
        }
    }

};

void T040() {
    vector<int> v;
    vector<vector<int>>Res;
    Solution s;
    v = {10,1,2,7,6,1,5 };
    Res = s.combinationSum2(v, 8);
    for (auto &a : Res) {
        for (auto b : a)
            cout << b << "  ";
        cout << endl;
    }
    cout << endl << "*********" << endl;
    /*v = { 2, 3,5 };
    Res = s.combinationSum(v, 8);
    for (auto &a : Res) {
        for (auto b : a)
            cout << b << "  ";
        cout << endl;
    }
    cout << endl << "*********" << endl;
*/


}