LeetCode Combination Sum II DFS

思想：

与上一题相比多了previous，即如果前面的这一层已经试验过了该数，就跳过它。

class Solution {
public:
    //Combination Sum II
    void dfs(vector<int> num, int target, int pos, vector<int> &value, vector<vector<int>> &res) {
        if(target == 0) {
            res.push_back(value);
            return;
        }
        int previous = -1;
        for(int i = pos; i < num.size(); i++) {
            if(target < num[i]) {
                return;
            }
            if(previous == num[i]) continue;
            value.push_back(num[i]);
            previous = num[i];
            dfs(num,target-num[i],i+1,value,res);
            value.pop_back();
        }
    }
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        sort(num.begin(),num.end());
        vector<vector<int>> res;
        vector<int> value;
        dfs(num,target,0,value,res);
        return res;
    }
};