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.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• 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] 

这道题跟之前那道 Combination Sum 组合之和 本质没有区别，只需要改动一点点即可，之前那道题给定数组中的数字可以重复使用，而这道题不能重复使用，只需要在之前的基础上修改两个地方即可，首先在递归的for循环里加上if (i > start && num[i] == num[i – 1]) continue; 这样可以防止res中出现重复项，然后就在递归调用combinationSum2DFS里面的参数换成i+1，这样就不会重复使用数组中的数字了，代码如下：

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        vector<vector<int> > res;
        vector<int> out;
        sort(num.begin(), num.end());
        combinationSum2DFS(num, target, 0, out, res);
        return res;
    }
    void combinationSum2DFS(vector<int> &num, int target, int start, vector<int> &out, vector<vector<int> > &res) {
        if (target < 0) return;
        else if (target == 0) res.push_back(out);
        else {
            for (int i = start; i < num.size(); ++i) {
                if (i > start && num[i] == num[i - 1]) continue;
                out.push_back(num[i]);
                combinationSum2DFS(num, target - num[i], i + 1, out, res);
                out.pop_back();
            }
        }
    }
};

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Combination Sum III

Combination Sum