Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.

Example：

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]

Note：

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

解释下题目：

就是找出所有能加起来等于target的组合，不可以有重复

1. 递归

实际耗时：18ms

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> lists = new ArrayList<>();
        Arrays.sort(candidates);
        List<Integer> list = new ArrayList<>();
        getResult(lists, list, candidates, target, -1);
        return lists;
    }

    public void getResult(List<List<Integer>> lists, List list, int[] candidates, int target, int start) {
        if (target < 0) {
            return;
        } else if (0 == target) {
            if (!lists.contains(new ArrayList<>(list))) {
                lists.add(new ArrayList<>(list));
                return;
            }
            return;
        } else {
            for (int i = start + 1; i < candidates.length; i++) {
                if (i > start + 1 && candidates[i] == candidates[i - 1]) {
                    //去重操作
                    continue;
                }
                list.add(candidates[i]);
                getResult(lists, list, candidates, target - candidates[i], i);
                list.remove(list.size() - 1);
            }
        }
    }

踩过的坑：{2,2,2} target = 2

  思路和之前的039一样，只是自己不能多次使用了，那么就进行去重操作呗。

时间复杂度O(2^n) 也很复杂了

空间复杂度O(1)