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

 1 package com.Liuyt;
 2 import java.io.FileInputStream;
 3 import java.io.FileNotFoundException;
 4 import java.util.ArrayList;
 5 import java.util.List;
 6 import java.util.Scanner;
 7 
 8 public class Main {
 9     public static List<String> list = new ArrayList<String>();
10     public static void permu(char[] str) {
11         if (str == null) {
12             return;
13         }
14         permu(str, 0);
15     }
16 
17     private static void permu(char[] str, int begin) {
18         if (begin == str.length) {
19             list.add(String.valueOf(str));
20         } else {
21             for (int i = begin; i < str.length; i++) {
22                 // 首次for循环时候i=begin，即a,b,c分别和自身交换
23                 char temp = str[begin];
24                 str[begin] = str[i];
25                 str[i] = temp;
26 
27                 permu(str, begin + 1);
28                 // 采用递归调用，每次begin+1后 带入新的递归
29                 /* 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！ */
30                 temp = str[begin];
31                 str[begin] = str[i];
32                 str[i] = temp;
33             }
34         }
35     }
36 
37     public static void main(String[] args) throws FileNotFoundException {
38         Scanner sc = new Scanner(System.in);
39 
40         String input = sc.nextLine();
41         String s1 = input.split(" ")[0];
42         String s2 = input.split(" ")[1];
43 
44         if (s2.length() < s1.length() || s2 == null){
45             System.out.println("false");
46             return;
47         }else if(s1 == null && s2 == null){
48             System.out.println("true");
49         }
50         char[] s1_char = s1.toCharArray();
51         permu(s1_char);
52 
53         for (int i = 0 ;i < list.size();i++){
54             System.out.println(list.get(i));
55             if (s2.contains(list.get(i))){
56                 System.out.println("true");
57                 return;
58             }
59         }
60         System.out.println("false");
61     }
62 }