[LeetCode]--39. 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.
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]
]

我想好半天也没把递归搞清楚，在网上看了一个比较好的算法，跟着学了一遍，debug跟着递归走一走，感觉拨开迷雾一样。

public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (candidates.length == 0)
            return res;
        List<Integer> temp = new ArrayList<Integer>();
        Arrays.sort(candidates);
        findSum(candidates, target, 0, 0, temp, res);
        return res;
    }

    private void findSum(int[] candidates, int target, int sum, int level,
            List<Integer> temp, List<List<Integer>> res) {
        if (sum == target) {
            res.add(new ArrayList<Integer>(temp));
            return;
        } else if (sum > target)
            return;
        else
            for (int i = level; i < candidates.length; i++) {
                temp.add(candidates[i]);
                findSum(candidates, target, sum + candidates[i], i, temp, res);
                temp.remove(temp.size() - 1);
            }
    }