Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

78. 子集 #33

Open
webVueBlog opened this issue Sep 4, 2022 · 0 comments
Open

78. 子集 #33

webVueBlog opened this issue Sep 4, 2022 · 0 comments

Comments

@webVueBlog
Copy link
Owner

78. 子集

Description

Difficulty: 中等

Related Topics: 位运算, 数组, 回溯

给你一个整数数组 nums ,数组中的元素 互不相同 。返回该数组所有可能的子集(幂集)。

解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。

示例 1:

输入:nums = [1,2,3]
输出:[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

示例 2:

输入:nums = [0]
输出:[[],[0]]

提示:

  • 1 <= nums.length <= 10
  • -10 <= nums[i] <= 10
  • nums 中的所有元素 互不相同

Solution

Language: JavaScript

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
// 递归法实现子集枚举
// var subsets = function (nums) {
//     let result = []
//     let path = []
//     function backtracking(startIndex) {
//         result.push(path.slice())
//         for (let i = startIndex; i < nums.length; i++) {
//             path.push(nums[i])
//             backtracking(i + 1)
//             path.pop()
//         }
//     }
//     backtracking(0)
//     return result
// }

// DFS回溯思路
var subsets = function (nums) {
    const res = []
    const dfs = (index, list) => {
        res.push(list.slice())
        for (let i = index; i < nums.length; i++) {
            list.push(nums[i])
            dfs(i+1, list)
            list.pop()
        }
    }
    dfs(0, [])
    return res
}
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Labels
None yet
Projects
None yet
Development

No branches or pull requests

1 participant