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]
]

 

解法：

和Combination Sum类似，有两个地方需要注意，修改递归中index的值为index+1，这样避免重复找数字。

第二个地方是重复字，比如1,7和7,1。所以先对candidates排序，然后在remove结果的时候跳过后面与其相同的数字。

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        Arrays.sort(candidates);
        findAns(res,tmp,target,candidates,0);
        return res;
    }
    
    public void findAns(List<List<Integer>> res, List<Integer> tmp, int target, int[] candidates, int index){
        if(target == 0){
            List<Integer> fl = new ArrayList<Integer>(tmp);
            res.add(fl);
        }else if(target<0){
            return;
        }else {
            for(int i=index;i<candidates.length;i++){
                tmp.add(candidates[i]);
                findAns(res,tmp,target-candidates[i],candidates,i+1);
                tmp.remove(tmp.size()-1);
                while(i<candidates.length-1 && candidates[i]==candidates[i+1]){
                    i++;
                }
            }
        }
    }
}