算法题：字符串s1,s2，判断s1的任意排列是否是s2的子串，返回true或false

package com.Liuyt;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static List<String> list = new ArrayList<String>();
    public static void permu(char[] str) {
        if (str == null) {
            return;
        }
        permu(str, 0);
    }

    private static void permu(char[] str, int begin) {
        if (begin == str.length) {
            list.add(String.valueOf(str));
        } else {
            for (int i = begin; i < str.length; i++) {
                // 首次for循环时候i=begin，即a,b,c分别和自身交换
                char temp = str[begin];
                str[begin] = str[i];
                str[i] = temp;

                permu(str, begin + 1);
                // 采用递归调用，每次begin+1后 带入新的递归
                /* 交换一遍后再交换一次，能够保证最后的到的还是原数组，好办法！ */
                temp = str[begin];
                str[begin] = str[i];
                str[i] = temp;
            }
        }
    }

    public static void main(String[] args) throws FileNotFoundException {
        Scanner sc = new Scanner(System.in);

        String input = sc.nextLine();
        String s1 = input.split(" ")[0];
        String s2 = input.split(" ")[1];

        if (s2.length() < s1.length() || s2 == null){
            System.out.println("false");
            return;
        }else if(s1 == null && s2 == null){
            System.out.println("true");
        }
        char[] s1_char = s1.toCharArray();
        permu(s1_char);

        for (int i = 0 ;i < list.size();i++){
            System.out.println(list.get(i));
            if (s2.contains(list.get(i))){
                System.out.println("true");
                return;
            }
        }
        System.out.println("false");
    }
}

示例1：

ab dshfsjbaooo

true

示例2：

ab dsjklacbsoo

false