排列数字（dfs)

#include <iostream>
using namespace std;
const int N=10;

int path[N];
bool st[N];

void dfs( int u)
{
	if(u==n)//这个是判断是否走到了终点
	{
		for(int i=0;i<n;i++) cout<<path[i];
		cout<<endl;
		return ; 
	}
	for(int i=1;i<=n;i++)
	{
		if(!st[i])//看是否走过
		{
			path[u]=i;
			st[i]=ture;
			dfs(u+1);
			st[i]=false;//回溯
		}
	}
} 

int main()
{
	cin>>n;
	dfs(0);
	return 0;
}

eg输入3，输出123 132…对1到3 的数进行全排列共有六种；
java：

package org.example;

import java.util.Scanner;

public class Permutations {
    private static final int N = 10;
    private static int[] path = new int[N];
    private static boolean[] st = new boolean[N];

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        dfs(n, 0);
    }

    private static void dfs(int n, int u) {
        if (u == n) {
            for (int i = 0; i < n; i++) {
                System.out.print(path[i] + " ");
            }
            System.out.println();
            return;
        }
        for (int i = 1; i <= n; i++) {
            if (!st[i]) {
                path[u] = i;
                st[i] = true;
                dfs(n, u + 1);
                st[i] = false; // 回溯
            }
        }
    }
}