comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3437.Permutations%20III/README.md	数组 回溯

3437. Permutations III 🔒

English Version

题目描述

Given an integer n, an alternating permutation is a permutation of the first n positive integers such that no two adjacent elements are both odd or both even.

Return all such alternating permutations sorted in lexicographical order.

 

Example 1:

Input: n = 4

Output: [[1,2,3,4],[1,4,3,2],[2,1,4,3],[2,3,4,1],[3,2,1,4],[3,4,1,2],[4,1,2,3],[4,3,2,1]]

Example 2:

Input: n = 2

Output: [[1,2],[2,1]]

Example 3:

Input: n = 3

Output: [[1,2,3],[3,2,1]]

 

Constraints:

• 1 <= n <= 10

解法

方法一：回溯

我们设计一个函数 $\textit{dfs}(i)$，表示当前要填第 $i$ 个位置的数，位置编号从 $0$ 开始。

在 $\textit{dfs}(i)$ 中，如果 $i \geq n$，说明所有位置都已经填完，将当前排列加入答案数组中。

否则，我们枚举当前位置可以填的数 $j$，如果 $j$ 没有被使用过，并且 $j$ 和当前排列的最后一个数不同奇偶性，我们就可以将 $j$ 放在当前位置，继续递归填下一个位置。

时间复杂度 $O(n \times n!)$，空间复杂度 $O(n)$。其中 $n$ 为排列的长度。

Python3

class Solution:
    def permute(self, n: int) -> List[List[int]]:
        def dfs(i: int) -> None:
            if i >= n:
                ans.append(t[:])
                return
            for j in range(1, n + 1):
                if not vis[j] and (i == 0 or t[-1] % 2 != j % 2):
                    t.append(j)
                    vis[j] = True
                    dfs(i + 1)
                    vis[j] = False
                    t.pop()

        ans = []
        t = []
        vis = [False] * (n + 1)
        dfs(0)
        return ans

Java

class Solution {
    private List<int[]> ans = new ArrayList<>();
    private boolean[] vis;
    private int[] t;
    private int n;

    public int[][] permute(int n) {
        this.n = n;
        t = new int[n];
        vis = new boolean[n + 1];
        dfs(0);
        return ans.toArray(new int[0][]);
    }

    private void dfs(int i) {
        if (i >= n) {
            ans.add(t.clone());
            return;
        }
        for (int j = 1; j <= n; ++j) {
            if (!vis[j] && (i == 0 || t[i - 1] % 2 != j % 2)) {
                vis[j] = true;
                t[i] = j;
                dfs(i + 1);
                vis[j] = false;
            }
        }
    }
}

C++

class Solution {
public:
    vector<vector<int>> permute(int n) {
        vector<vector<int>> ans;
        vector<bool> vis(n);
        vector<int> t;
        auto dfs = [&](this auto&& dfs, int i) -> void {
            if (i >= n) {
                ans.push_back(t);
                return;
            }
            for (int j = 1; j <= n; ++j) {
                if (!vis[j] && (i == 0 || t[i - 1] % 2 != j % 2)) {
                    vis[j] = true;
                    t.push_back(j);
                    dfs(i + 1);
                    t.pop_back();
                    vis[j] = false;
                }
            }
        };
        dfs(0);
        return ans;
    }
};

Go

func permute(n int) (ans [][]int) {
	vis := make([]bool, n+1)
	t := make([]int, n)
	var dfs func(i int)
	dfs = func(i int) {
		if i >= n {
			ans = append(ans, slices.Clone(t))
			return
		}
		for j := 1; j <= n; j++ {
			if !vis[j] && (i == 0 || t[i-1]%2 != j%2) {
				vis[j] = true
				t[i] = j
				dfs(i + 1)
				vis[j] = false
			}
		}
	}
	dfs(0)
	return
}

TypeScript

function permute(n: number): number[][] {
    const ans: number[][] = [];
    const vis: boolean[] = Array(n).fill(false);
    const t: number[] = Array(n).fill(0);
    const dfs = (i: number) => {
        if (i >= n) {
            ans.push([...t]);
            return;
        }
        for (let j = 1; j <= n; ++j) {
            if (!vis[j] && (i === 0 || t[i - 1] % 2 !== j % 2)) {
                vis[j] = true;
                t[i] = j;
                dfs(i + 1);
                vis[j] = false;
            }
        }
    };
    dfs(0);
    return ans;
}