[编程题]: 给定一个数字n，求[1, n]这n个数字的排列组合有多少个（阿里2021校招笔试）

给定一个数字n，求[1, n]这n个数字的排列组合有多少个
条件：相邻的两个数之间的绝对值大于1
按字典序输出

示例

输入
4
输出
2 4 1 3
3 1 4 2

解题思路
采用DFS或者回溯

代码

Java

import java.util.*;

public class Main{

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        boolean[] visited = new boolean[n + 1];
        helper(n, visited, 0, new ArrayList<>());
    }

    public static void helper(int n, boolean[] visited, int count, List<Integer> res) {
        if (count == n) {
            for (int i = 0; i < n; i++) {
                System.out.print(res.get(i));
                if (i < n - 1) System.out.print(" ");
                else System.out.println();
            }
        } else {
            for (int i = 1; i <= n; i++) {
                if (!visited[i]) {
                    if (res.isEmpty() || Math.abs(res.get(res.size() - 1) - i) > 1) {
                        visited[i] = true;
                        res.add(i);
                        helper(n, visited, count + 1, res);
                        res.remove(res.size() - 1);
                        visited[i] = false;
                    }
                }
            }
        }
    }
}

输出

4
2 4 1 3
3 1 4 2

5
1 3 5 2 4
1 4 2 5 3
2 4 1 3 5
2 4 1 5 3
2 5 3 1 4
3 1 4 2 5
3 1 5 2 4
3 5 1 4 2
3 5 2 4 1
4 1 3 5 2
4 2 5 1 3
4 2 5 3 1
5 2 4 1 3
5 3 1 4 2

C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

void helper(int n, vector<bool>& visited, int count, vector<int> res) {
	if (count == n) {
		for (int i = 0; i < n; i++) {
			cout << res[i];
			if (i < n - 1) cout << " ";
			else cout << endl;
		}
	}
	else {
		for (int i = 1; i <= n; i++) {
			if (!visited[i]) {
				if (res.empty() || abs(res[res.size() - 1] - i) > 1) {
					res.push_back(i);
					visited[i] = true;
					helper(n, visited, count + 1, res);
					res.pop_back();
					visited[i] = false;
				}
			}
		}
	}
}

int main() {
	int n;
	cin >> n;
	vector<bool> visted(n + 1, false);
	vector<int> res;
	helper(n, visted, 0, res);
	return 0;
}

输出

4
2 4 1 3
3 1 4 2

5
1 3 5 2 4
1 4 2 5 3
2 4 1 3 5
2 4 1 5 3
2 5 3 1 4
3 1 4 2 5
3 1 5 2 4
3 5 1 4 2
3 5 2 4 1
4 1 3 5 2
4 2 5 1 3
4 2 5 3 1
5 2 4 1 3
5 3 1 4 2