40. Combination Sum II

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 target8
,A solution set is:
[ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]

public class Solution {
    private List<List<Integer>> ans = new ArrayList<List<Integer>>();
    
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        if(candidates.length==0||target<=0)
           return null;
        List<Integer> res = new ArrayList<Integer>();
        Arrays.sort(candidates);
        fun(candidates,0,target,res);
        return ans;
          
    }
    private void fun(int[] candidates,int start,int target,List<Integer> curr)
    {
        if(target == 0)
        {  
            List<Integer> list = new ArrayList<Integer>(curr);//若没有这一句，直接用ans.add(curr)的话，curr每次是变化的。
            ans.add(list);
        }
        else
        {
            for(int i = start;i<candidates.length;i++)
            {
                if(i!=start&&candidates[i]==candidates[i-1])  //去除重复
                    continue;
                if(target>=candidates[i])
                {
                    curr.add(candidates[i]);
                    fun(candidates,i+1,target-candidates[i],curr); //下次递归不能从当前位置开始
                    curr.remove(curr.size()-1);
                }
            }
        }
    }
}