给定一组数和一个目标值，返回和为目标值的集合（集合中的元素可重复）

以下为原题：

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where
the candidate numbers sums to target.The same repeated number may be chosen from candidates unlimited number of times.Note:All numbers (including
target) will be positive integers.The solution set must not contain duplicate combinations.
Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]
Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

思路：

1.以target<=0为基准：若target<0,return;若target=0,add,return。

2.每一层递归都遍历给定的数组，遍历的起始位置是上一层递归正在遍历的位置。如果target>0,则在下一层递归中target减去当前递归层正被遍历的元素。

代码如下：

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if(candidates==null||candidates.length==0)
            return new ArrayList<>();
        List<List<Integer>> allRes = new ArrayList<>();
        combinationSumDetail(candidates,target,new ArrayList<>(),allRes,0);
        return allRes;
    }

    private void combinationSumDetail(int[] candidates, int target, List<Integer> subRes, List<List<Integer>> allRes, int start){
        if(target<0)
            return;
        if(target==0){
            allRes.add(new ArrayList<>(subRes));
            return;
        }
        for(int i=start;i<candidates.length;++i){
            subRes.add(candidates[i]);
            combinationSumDetail(candidates,target-candidates[i],subRes,allRes,i);
            subRes.remove(subRes.size()-1);
        }
    }