题目：

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]
]

分析：

本题是LeetCode 39. Combination Sum 组合总和 (C++/Java)的扩展，我们还是利用39题中的方法来做这道题。

只不过本题所给的数字有重复，且要求最后的结果中不能有重复的，例如例子1，有两个1，他们都可以和7组合成8，如果按照39题搜索的方法就会产生重复结果。

其中一个办法就是将搜索生成的结果加入到set中，这样集合类会自动帮我们去重。

此外我们还可以在每一轮搜索时，当发现有相同元素时，直接跳过本次搜素，这样就不会产生重复的结果了。

程序：

C++

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

Java

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        LinkedList<Integer> curr = new LinkedList<>();
        Arrays.sort(candidates);
        dfs(candidates, target, 0, res, curr);
        return res;
    }
    private void dfs(int[] candidates, int target, int index, List<List<Integer>> res, LinkedList<Integer> curr){
        if(target == 0){
            res.add(new LinkedList<>(curr));
            return;
        }
        for(int i = index; i < candidates.length; ++i){
            if(candidates[i] > target)
                return;
            if(i > index && candidates[i] == candidates[i-1])
                continue;
            curr.addLast(candidates[i]);
            dfs(candidates, target-candidates[i], i+1, res, curr);
            curr.removeLast();
        }
    }
}