题目描述

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.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
Example 1:

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]
]
Example 2:

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

算法

时间复杂度 我已经算不清楚了,感觉是O(2^n)

参考文献

Java 代码

class Solution {

    List<List<Integer>> res = new ArrayList<>();

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        int n = candidates.length;
        if (n == 0 || target == 0) {
            return res;
        }

        Arrays.sort(candidates);

        dfs(candidates, target, 0, new ArrayList<Integer>());

        return res;
    }

    private void dfs(int[] candidates, int target, int start, List<Integer> path) {
        if (target < 0) {
            return;
        }
        if (target == 0) {
            res.add(new ArrayList<>(path));
            return;
        }

        for (int i = start; i < candidates.length; i++) {
            // 剪枝1，因为排序过，直接break, 后面不用管了
            if (target < candidates[i]) {
                break;
            }
            // 剪枝2，过滤重复解,注意i > start本次枚举的起点,如果出现1,2,2,2,5,6,9,10,需要去掉第2，3个2
            if (i > start && candidates[i] == candidates[i - 1]) {
                continue;
            }

            path.add(candidates[i]);
            dfs(candidates, target - candidates[i], i + 1, path);
            path.remove(path.size() - 1);
        }
    }

}