leetcode-39

给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。

说明：

所有数字（包括 target）都是正整数。
解集不能包含重复的组合。 
示例 1:

输入: candidates = [2,3,6,7], target = 7,
所求解集为:
[
[7],
[2,2,3]
]
示例 2:

输入: candidates = [2,3,5], target = 8,
所求解集为:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/combination-sum
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这道题就是一个回溯的思路。

可以看https://leetcode-cn.com/problems/combination-sum/solution/hui-su-suan-fa-jian-zhi-python-dai-ma-java-dai-m-2/

public class Solution {
    private List<List<Integer>> res = new ArrayList<>();
    private int[] candidate;
    private int len;

    private List<List<Integer>> combinationSum(int[] candidate, int target) {
        int len = candidate.length;
        if (len == 0) {
            return res;
        }
        Arrays.sort(candidate);
        this.len = len;
        this.candidate = candidate;
        findCombinationSum(target, 0, new Stack<Integer>());
        return res;
    }

    private void findCombinationSum(int residue, int start, Stack<Integer> pre) {
        if (residue == 0) {
            res.add(new ArrayList<Integer>(pre));
            return;
        }
        for (int i = start; i < len && residue - candidate[i] >= 0; i++) {
            pre.push(candidate[i]);
            findCombinationSum(residue - candidate[i], i, pre);
            pre.pop();
        }
    }

    public static void main(String[] args) {
        int[] candidates = {3, 4, 6, 7};
        int target = 7;
        Solution solution = new Solution();
        List<List<Integer>> combinationSum = solution.combinationSum(candidates, target);
        System.out.println(combinationSum);

    }
}

end