Leetcode: Combination Sum

Given a set of candidate numbers (C) 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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

典型的recursion方法，找符合要求的path，存入result的ArrayList中。所以方法还是建立一个ArrayList<ArrayList<Integer>> result, 建立一个ArrayList<Integer> path，用recursion当找到符合条件的path时，存入result中。

我在做这道题时遇到了一个问题：添加 path 进入 result 中时，需要这样res.add(new ArrayList<Integer>(path)); 如果直接res.add(path); 会出错

比如我遇到的错误是：Input:[1], 1    Output:[[]]     Expected:[[1]]，没能够把path: [1] 添加到res里面去，没有成功。（具体我现在也不知道为什么）

第二遍做法：

public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> item = new ArrayList<Integer>();
        Arrays.sort(candidates);
        helper(res, item, candidates, target, 0);
        return res;
    }
    
    public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> item, int[] candidates, int remain, int start) {
        if (remain < 0) return;
        if (remain == 0) {
            res.add(new ArrayList<Integer>(item));
            return;
        }
        for (int i=start; i<candidates.length; i++) {
            if (i>start && candidates[i] == candidates[i-1]) continue;
            item.add(candidates[i]);
            helper(res, item, candidates, remain-candidates[i], i);
            item.remove(item.size()-1);
        }
    }
}