回文字符串问题

一、判断一个字符串是否为回文字符串（aba）

    public static boolean is_hwzfs(String s, int i, int j) {
        if (i > j) return false;
        for (int k = i; k <= j; ++k) {
            if (s.charAt(k) != s.charAt(j + i - k)) {
                return false;
            }
        }
        return true;
    }

二、求一个字符串的最长回文字符串长度并输出

1）暴力枚举

 1 public static boolean is_hwzfs(char[] s, int i, int j) {
 2         if (i > j) return false;
 3         for (int k = i; k <= j; ++k) {
 4             if (s[k] != s[j + i - k]) {
 5                 return false;
 6             }
 7         }
 8         return true;
 9     }
10 
11     public static void main(String[] args) {
12 
13         Scanner cin = new Scanner(System.in);
14         String s = cin.nextLine();
15 
16         int len = s.length();
17         int max = 1;
18         int x = 0, y = 0;
19         int[] p = new int[len];
20         char[] ans = new char[len];
21         int index = -1;
22         for (int i = 0; i < len; ++i) {
23             char ch = s.charAt(i);
24             if (Character.isLetter(ch)) {
25                 p[++index] = i;
26                 ans[index] = Character.toLowerCase(ch);
27             }
28         }
29 
30 
31         for (int i = 0; i <= index; ++i) {
32             for (int j = i; j <= index; ++j) {
33                 if (is_hwzfs(ans, i, j)) {
34                     if (max < (j - i + 1)) {
35                         max = j - i + 1;
36                         x = p[i];
37                         y = p[j];
38                     }
39                 }
40             }
41         }
42         System.out.println("max= " + max);
43         for (int i = x; i <= y; ++i) {
44             System.out.print(s.charAt(i));
45         }
46         System.out.println();
47     }

 

2）确定字符串中心，向两边扩展枚举

  1. 长度为奇数:  判断 s[i-j]与s[i+j]  (2*j+1)
  2. 长度为偶数：判断 s[i-j]与 s[i+j+1] (2*(j+1))

注意 i为字符串中心，j为向两边扩展的长度

法一：

法二：

    public static void main(String[] args) {

        Scanner cin = new Scanner(System.in);
        String s = cin.nextLine();

        int len = s.length();
        int max = 1;
        int x = 0, y = 0;
        int[] p = new int[len];
        char[] ans = new char[len];
        int index = -1;
        for (int i = 0; i < len; ++i) {
            char ch = s.charAt(i);
            if (Character.isLetter(ch)) {
                p[++index] = i;
                ans[index] = Character.toLowerCase(ch);
            }
        }

        for (int i = 0; i <= index; ++i) {

            for (int j = 0; (i - j > 0) && (i + j <= index); ++j) {
                if (ans[i - j] != ans[i + j]) break;
                if ((2 * j + 1) > max) {
                    x = p[i - j];
                    y = p[i + j];
                    max = 2 * j + 1;
                }
            }

            for (int j = 0; (i - j > 0) && (i + j + 1 <= index); ++j) {
                if (ans[i - j] != ans[i + j + 1]) break;
                if (2 * (j + 1) > max) {
                    x = p[i - j];
                    y = p[i + j + 1];
                    max = 2 * (j + 1);
                }
            }
        }

        System.out.println("max= "+max);
        for (int i = x; i <= y; ++i) {
            System.out.print(s.charAt(i));
        }
        System.out.println();


    }

 

posted @ 2016-09-03 13:45  奋斗的珞珞  阅读(233)  评论(0)  编辑  收藏  举报