Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
For example, given candidate set [2, 3, 6, 7] and target7,A solution set is:
[ [7], [2, 2, 3] ]

public class Solution {
    private List<List<Integer>> ans = new ArrayList<List<Integer>>();
    
    public List<List<Integer>> combinationSum(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);//若沒有這一句，直接用ans.add(curr)的話，curr每次是變化的。
            ans.add(list);
        }
        else
        {
            for(int i = start;i<candidates.length;i++)
            {
                if(target>=candidates[i])
                {
                    curr.add(candidates[i]);
                    fun(candidates,i,target-candidates[i],curr);
                    curr.remove(curr.size()-1);
                }
            }
        }
    }
}