【LeetCode算法练习（C++）】Combination Sum II

题目：
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

链接：Combination Sum II

解法：与Combination Sum类似，不同的是递归时需要排除已经选择的数字，答案用set存以过滤重复的数组。时间O(n^2)

class Solution {public:    vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {        set<vector<int> > ans;        vector<int> out;        sort(candidates.begin(), candidates.end());        combinationSumDFS(candidates, target, 0, out, ans);        return vector<vector<int> >(ans.begin(), ans.end());    }    void combinationSumDFS(vector<int> &candidates, int target, int start, vector<int> &out, set<vector<int> > &res) {        if (target < 0) return;        else if (target == 0) res.insert(out);        else             for (int i = start; i < candidates.size(); ++i) {                int tmp = candidates[i];                out.push_back(tmp);                candidates.erase(candidates.begin() + i);                combinationSumDFS(candidates, target - tmp, i, out, res);                candidates.insert(candidates.begin() + i, out[out.size() - 1]);                out.pop_back();            }    }};

Runtime: 93 ms