1 Star 0 Fork 332

志强 / leetcode

forked from doocs / leetcode 
加入 Gitee
与超过 1200万 开发者一起发现、参与优秀开源项目,私有仓库也完全免费 :)
免费加入
克隆/下载
README_EN.md 2.09 KB
一键复制 编辑 原始数据 按行查看 历史
ylb 提交于 2021-04-21 23:12 . feat: update leetcode problems description

90. Subsets II

中文文档

Description

Given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

 

Example 1:

Input: nums = [1,2,2]
Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]

Example 2:

Input: nums = [0]
Output: [[],[0]]

 

Constraints:

  • 1 <= nums.length <= 10
  • -10 <= nums[i] <= 10

Solutions

Python3

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        def dfs(nums, i, res, path):
            res.append(copy.deepcopy(path))
            for j in range(i, len(nums)):
                if j != i and nums[j] == nums[j - 1]:
                    continue
                path.append(nums[j])
                dfs(nums, j + 1, res, path)
                path.pop()
        res, path = [], []
        nums.sort()
        dfs(nums, 0, res, path)
        return res

Java

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<Integer> path = new ArrayList<>();
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);
        dfs(nums, 0, res, path);
        return res;
    }

    private void dfs(int[] nums, int i, List<List<Integer>> res, List<Integer> path) {
        res.add(new ArrayList<>(path));
        for (int j = i; j < nums.length; ++j) {
            if (j != i && nums[j] == nums[j - 1]) {
                continue;
            }
            path.add(nums[j]);
            dfs(nums, i + 1, res, path);
            path.remove(path.size() - 1);
        }
    }
}

...

Java
1
https://gitee.com/zzq3708/leetcode.git
git@gitee.com:zzq3708/leetcode.git
zzq3708
leetcode
leetcode
main

搜索帮助

53164aa7 5694891 3bd8fe86 5694891