7.17dfs例题：困难的串

 1 import java.util.Scanner;
 2 
 3 public class Seven_17dfs困难的串 {
 4     static int count = 0;
 5 
 6     private static void dfs(int l, int n, String prefix) {
 7 
 8         // 尝试在prefix后追加一个字符
 9         for (char i = ‘A‘; i < ‘A‘ + l; i++) {
10             if (isHard(prefix, i)) {// 是困难的串,就组合起来输出
11                 String x = prefix + i;
12                 System.out.println(x);
13                 count++;// 计数
14                 if (count == n)
15                     System.exit(0);
16 
17                 dfs(l, n, x);
18             }
19         }
20     }
21 
22     /**
23      * 判断prefix+i是否一个困难的串 1.遍历所有的长度为偶数的子串,看是否对称 2.prefix是一个困难的串 ABACA i
24      */
25     private static boolean isHard(String prefix, char i) {
26         int wide = 0;// 截取的宽度
27         for (int j = prefix.length() - 1; j >= 0; j -= 2) {
28             final String s1 = prefix.substring(j, j + wide + 1);
29             final String s2 = prefix.substring(j + wide + 1) + i;
30             if (s1.equals(s2))
31                 return false;
32             wide++;
33         }
34         return true;
35     }
36 
37     public static void main(String[] args) {
38         Scanner in = new Scanner(System.in);
39         int l = in.nextInt();
40         int n = in.nextInt();
41 
42         dfs(l, n, "");
43     }
44 }