1. 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]
]

2. 思路

和上一题相比主要是两点，第一同一个数字不能重复，第二是当有重复候选时，同一个组合算一种。
直接复用上一次的代码，区别有两点：

1. 递归时不是从当前位置开始限制，而是从下一个位置开始限制，也就是不可以使用当前这个已经使用过了的数字

2. 每次在取值上对同一个targe的当前场景不和上一个重复。

3. 代码

耗时：23ms

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        return combinationSum(candidates, target, 0);
    }
    
    vector<vector<int>> combinationSum(vector<int>& candidates, int target, int beg) {
        vector<vector<int>> ret;
        if (target <= 0) {return ret;}
        for (int i = beg; i < candidates.size(); i++) {
            if (i != beg && candidates[i] == candidates[i-1]) {continue;} // 避免重复
            int iv = candidates[i];
            if (iv == target) {
                vector<int> one;
                one.push_back(iv);
                ret.push_back(one);
            }
            
            vector<vector<int>> suf = combinationSum(candidates, target-iv, i + 1);
            for (int j = 0; j < suf.size(); j++) {
                suf[j].push_back(iv);
                ret.push_back(suf[j]);
            }
        }
        return ret;
    }
};