LeetCoder_____ 组合总和 II(40)

1. 思路

和前一个题目类似，只不过这里要求每个数只能用一次。
所以这里我们需要先将数组排序。然后dfs递归的时候保证不选取重复的数字即可。

2.代码

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> ret;
        vector<int> temp;
        sort(candidates.begin(), candidates.end());
        dfs(ret, 0, 0, 0, temp, candidates, target);
        return ret;
    }

    void dfs(vector<vector<int>>& ret, int sum, int pos, int start, vector<int>& temp, vector<int>& candidates, int target)
    {   
        if(sum == target)
        {
            temp.resize(pos);
            ret.push_back(temp);
            return;
        }
        temp.resize(pos + 1);
        for(int i = start; i < candidates.size(); ++i)
        {
            if(sum + candidates[i] > target || 
            		i != start && candidates[i] == candidates[i-1]) //注意这个判断
                continue;
            temp[pos] = candidates[i];
            dfs(ret, sum + candidates[i], pos + 1, i + 1, temp, candidates, target);
        }
    }
};