Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates 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.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

  现在每个candidate只能用一次了，那不就在前一题的基础上从pos+1的位置开始找就好了？ 但是还有别的细节要考虑：candidate里是有一样的数的，但它们是不同的candidates。比如[10, 1a, 2, 7, 6, 1b, 5]里，[1a, 1b, 6]这种情况的确能被考虑进来，但[1a, 7], [1b, 7]都会被考虑进来，就出现duplicates了。解决办法是加个小条件，还是基于对原数组排序的基础上，如果nums里当前数和前一数相等（1b和1a），就在这层规避掉这个分支。也就是说，1a，1b能在不同层被加入到同一个combination里，但不能在同一层被加入到不同的combination里。   实现：

class Solution {
private:
    void dfs(vector<int>& candidates, int target, vector<vector<int>>& res, vector<int>& combination, int pos){
        if (target == 0){
            res.push_back(combination);
            return;
        }

        for (int i=pos; i<candidates.size(); i++){
            if (candidates[i] > target)
                return;
            if (i>pos && candidates[i] == candidates[i-1])
                continue;
            combination.push_back(candidates[i]);
            dfs(candidates, target-candidates[i], res, combination, i+1);
            combination.pop_back();
        }
    }
    
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        
        vector<vector<int>> res;
        vector<int> combination;
        sort(candidates.begin(), candidates.end());
        dfs(candidates, target, res, combination, 0);
        return res;
        
    } 
};