求{1,2,3}的全排列类的问题

最近遇到了一个题，求3个数全排列所组成的数字，想类似蒙特卡罗法，多次循环将所有排列的可能打出来，于是就有以下代码

package com.dong.test5;
import java.util.Arrays;
import java.util.Random;


/*
五、分析以下需求，并用代码实现：
		(1)打印由7，8，9三个数组成的三位数，要求该三位数中任意两位数字不能相同；
		(2)打印格式最后的三位数字以空格分隔，如789 798 879 897 978 987。
    注：要求使用StringBuilder来完成

* */
public class Test5 {
    public static void main(String[] args) {
        StringBuilder sb = new StringBuilder();
        String result ="";

        String[] strings = {"7","8","9"};

        for (int i = 0; i < 50; i++) {
            int index1 = new Random().nextInt(3);

            int index2 = new Random().nextInt(3);
            while(index1==index2){
                 index2 = new Random().nextInt(3);
            }

            int index3 = new Random().nextInt(3);
            while(index3 ==index1 || index3 == index2){
                index3 = new Random().nextInt(3);
            }

            result = strings[index1]+strings[index2]+strings[index3];
            if(sb.indexOf(result)<0){
                sb.append(result+" ");
            }

        }
        //将随机的输出序列排个序，变成稳定的输出结果

        String[] split = sb.toString().split(" ");
        int[] arr = new int[split.length];
        for (int i = 0; i < split.length; i++) {
            arr[i] = Integer.valueOf(split[i]);
        }
        Arrays.sort(arr);
        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i]+" ");
        }


    }
}

当然了，上面的写法还是太笨拙，抽空用递归的写法补充一下

package NewCodeAlgorithm;

import java.util.ArrayList;

/*
 * 打印一个字符串的全排列
 *
 * */
public class Print_All_Permutations {
    public static void main(String[] args) {
        String test = "abc";
        char[] c = test.toCharArray();
        ArrayList<String> list = new ArrayList<>();
        permutations(c, 0, list);
        System.out.println("++++++++++++++++++++");
        System.out.println(list);
    }

    public static void permutations(char[] c, int i, ArrayList<String> list) {
        if (i == c.length - 1) {
            if (!list.contains(String.valueOf(c))) {
                System.out.println(String.valueOf(c));
                list.add(String.valueOf(c));
            }
            return;
        }

        for (int j = i; j < c.length; j++) {
            swap(c, i, j);
            permutations(c, i + 1, list);
            //把字符数组复位，再去尝试另外一位交换
            swap(c, i, j);
        }

    }

    public static void swap(char[] c, int i, int j) {
        char temp = c[i];
        c[i] = c[j];
        c[j] = temp;
    }
}

其他递归的思路可以参考下文

https://blog.csdn.net/webpain/article/details/79601278