面试题 08.04. 幂集

English Version

题目描述

幂集。编写一种方法，返回某集合的所有子集。集合中不包含重复的元素。

说明：解集不能包含重复的子集。

示例:

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

解法

回溯法

Python3

Java

JavaScript

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsets = function (nums) {
    let prev = [];
    let res = [];
    dfs(nums, 0, prev, res);
    return res;
};

function dfs(nums, depth, prev, res) {
    res.push(prev.slice());
    for (let i = depth; i < nums.length; i++) {
        prev.push(nums[i]);
        depth++;
        dfs(nums, depth, prev, res);
        prev.pop();
    }
}

TypeScript

function subsets(nums: number[]): number[][] {
    const res = [[]];
    nums.forEach(num => {
        res.forEach(item => {
            res.push(item.concat(num));
        });
    });
    return res;
}

function subsets(nums: number[]): number[][] {
    const n = nums.length;
    const res = [];
    const list = [];
    const dfs = (i: number) => {
        if (i === n) {
            res.push([...list]);
            return;
        }
        list.push(nums[i]);
        dfs(i + 1);
        list.pop();
        dfs(i + 1);
    };
    dfs(0);
    return res;
}

Rust

impl Solution {
    pub fn subsets(nums: Vec<i32>) -> Vec<Vec<i32>> {
        let n = nums.len();
        let mut res: Vec<Vec<i32>> = vec![vec![]];
        for i in 0..n {
            for j in 0..res.len() {
                res.push(vec![..res[j].clone(), vec![nums[i]]].concat());
            }
        }
        res
    }
}

impl Solution {
    fn dfs(nums: &Vec<i32>, i: usize, res: &mut Vec<Vec<i32>>, list: &mut Vec<i32>) {
        if i == nums.len() {
            res.push(list.clone());
            return;
        }
        list.push(nums[i]);
        Self::dfs(nums, i + 1, res, list);
        list.pop();
        Self::dfs(nums, i + 1, res, list);
    }

    pub fn subsets(nums: Vec<i32>) -> Vec<Vec<i32>> {
        let mut res = vec![];
        Self::dfs(&nums, 0, &mut res, &mut vec![]);
        res
    }
}

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