3437. Permutations III 🔒

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Description

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:

Example 2:

Example 3:

 

Constraints:

• 1 <= n <= 10

Solutions

Solution 1: Backtracking

We design a function $\textit{dfs}(i)$, which represents filling the $i$-th position, with position indices starting from $0$.

In $\textit{dfs}(i)$, if $i \geq n$, it means all positions have been filled, and we add the current permutation to the answer array.

Otherwise, we enumerate the numbers $j$ that can be placed in the current position. If $j$ has not been used and $j$ has a different parity from the last number in the current permutation, we can place $j$ in the current position and continue to recursively fill the next position.

The time complexity is $O(n \times n!)$, and the space complexity is $O(n)$. Here, $n$ is the length of the permutation.

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;
}