40. 组合总和 II Combination Sum II

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.

Each number in candidates may only be used once in the combination.

Note: The solution set must not contain duplicate combinations.

Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

方法：

回溯加剪枝。先排序，当同层有相同的元素时，只需要取第一个结果即可。

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        int len = candidates.length;
        List<List<Integer>> res = new ArrayList<>();
        if (len == 0) {
            return res;
        }
        Arrays.sort(candidates);

        Deque<Integer> path = new ArrayDeque<>(len);
        dfs(candidates, len, 0, target, path, res);
        return res;
    }
    private void dfs(int[] candidates, int len, int begin, int target, Deque<Integer> path, List<List<Integer>> res) {
        if (target == 0) {
            res.add(new ArrayList<>(path));
            return;
        }
        for (int i = begin; i < len; i++) {
            if (target - candidates[i] < 0) {
                break;
            }
            
            if (i > begin && candidates[i] == candidates[i - 1]) {
                continue;
            }

            path.addLast(candidates[i]);
            
            dfs(candidates, len, i + 1, target - candidates[i], path, res);

            path.removeLast();
        }
    }