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剑指 Offer II 083. 没有重复元素集合的全排列
题目描述
给定一个不含重复数字的整数数组 nums ，返回其 所有可能的全排列 。可以 按任意顺序 返回答案。
示例 1：
输入：nums = [1,2,3] 输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
示例 2：
输入：nums = [0,1] 输出：[[0,1],[1,0]]
示例 3：
输入：nums = [1] 输出：[[1]]
提示：
1 <= nums.length <= 6-10 <= nums[i] <= 10nums中的所有整数 互不相同
注意：本题与主站 46 题相同：https://leetcode.cn/problems/permutations/
解法
方法一
Python3
class Solution:def permute(self, nums: List[int]) -> List[List[int]]:n=len(nums)res=[]path=[]def dfs(i):if len(path)==n:res.append(path[:])returnif len(path)>n:returnfor j in range(n):if nums[j] in path:continue # 也可以空间换时间used[j]path.append(nums[j])dfs(j)path.pop()dfs(0)return res
Java
class Solution {public List<List<Integer>> permute(int[] nums) {List<List<Integer>> res = new ArrayList<>();List<Integer> path = new ArrayList<>();int n = nums.length;boolean[] used = new boolean[n];dfs(0, n, nums, used, path, res);return res;}private void dfs(int u, int n, int[] nums, boolean[] used, List<Integer> path, List<List<Integer>> res) {if (u == n) {res.add(new ArrayList<>(path));return;}for (int i = 0; i < n; ++i) {if (!used[i]) {path.add(nums[i]);used[i] = true;dfs(u + 1, n, nums, used, path, res);used[i] = false;path.remove(path.size() - 1);}}}
}
C++
class Solution {
public:vector<vector<int>> permute(vector<int>& nums) {int n = nums.size();vector<vector<int>> res;vector<int> path(n, 0);vector<bool> used(n, false);dfs(0, n, nums, used, path, res);return res;}void dfs(int u, int n, vector<int>& nums, vector<bool>& used, vector<int>& path, vector<vector<int>>& res) {if (u == n) {res.emplace_back(path);return;}for (int i = 0; i < n; ++i) {if (!used[i]) {path[u] = nums[i];used[i] = true;dfs(u + 1, n, nums, used, path, res);used[i] = false;}}}
};
Go
func permute(nums []int) [][]int {n := len(nums)res := make([][]int, 0)path := make([]int, n)used := make([]bool, n)dfs(0, n, nums, used, path, &res)return res
}func dfs(u, n int, nums []int, used []bool, path []int, res *[][]int) {if u == n {t := make([]int, n)copy(t, path)*res = append(*res, t)return}for i := 0; i < n; i++ {if !used[i] {path[u] = nums[i]used[i] = truedfs(u+1, n, nums, used, path, res)used[i] = false}}
}
JavaScript
/*** @param {number[]} nums* @return {number[][]}*/
var permute = function (nums) {const n = nums.length;let res = [];let path = [];let used = new Array(n).fill(false);dfs(0, n, nums, used, path, res);return res;
};function dfs(u, n, nums, used, path, res) {if (u == n) {res.push(path.slice());return;}for (let i = 0; i < n; ++i) {if (!used[i]) {path.push(nums[i]);used[i] = true;dfs(u + 1, n, nums, used, path, res);used[i] = false;path.pop();}}
}
C#
using System.Collections.Generic;
using System.Linq;public class Solution {public IList<IList<int>> Permute(int[] nums) {var results = new List<IList<int>>();var temp = new List<int>();var visited = new bool[nums.Length];Search(nums, visited, temp, results);return results;}private void Search(int[] nums, bool[] visited, IList<int> temp, IList<IList<int>> results){int count = 0;for (var i = 0; i < nums.Length; ++i){if (visited[i]) continue;++count;temp.Add(nums[i]);visited[i] = true;Search(nums, visited, temp, results);temp.RemoveAt(temp.Count - 1);visited[i] = false;}if (count == 0 && temp.Any()){results.Add(new List<int>(temp));}}
}
Swift
class Solution {func permute(_ nums: [Int]) -> [[Int]] {var res = [[Int]]()var path = [Int]()var used = [Bool](repeating: false, count: nums.count)dfs(0, nums.count, nums, &used, &path, &res)return res}private func dfs(_ u: Int, _ n: Int, _ nums: [Int], _ used: inout [Bool], _ path: inout [Int], _ res: inout [[Int]]) {if u == n {res.append(path)return}for i in 0..<n {if !used[i] {path.append(nums[i])used[i] = truedfs(u + 1, n, nums, &used, &path, &res)used[i] = falsepath.removeLast()}}}
}
