Lintcode: Longest Common Substring 解题报告

最新推荐文章于 2025-12-08 09:18:56 发布

转载最新推荐文章于 2025-12-08 09:18:56 发布 · 137 阅读

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本文深入探讨了寻找两个字符串中最长公共子串的问题，通过动态规划方法详细解释了算法的实现过程，包括状态转移方程和代码示例。

Longest Common Substring

原题链接： http://lintcode.com/zh-cn/problem/longest-common-substring/#

Given two strings, find the longest common substring.

Return the length of it.

注意

The characters in substring should occur continiously in original string. This is different with subsequnce.

样例

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Dynamic Programming Longest Common Subsequence LintCode Copyright

SOLUTION 1:

DP:

D[i][j] 定义为：两个string的前i个和前j个字符串，尾部连到最后的最长子串。

D[i][j] =

1. i = 0 || j = 0 : 0

2. s1.char[i - 1] = s2.char[j - 1] ? D[i-1][j-1] + 1 : 0;

另外，创建一个max的缓存，不段更新即可。

 1 public class Solution {
 2     /**
 3      * @param A, B: Two string.
 4      * @return: the length of the longest common substring.
 5      */
 6     public int longestCommonSubstring(String A, String B) {
 7         // write your code here
 8         if (A == null || B == null) {
 9             return 0;
10         }
11         
12         int lenA = A.length();
13         int lenB = B.length();
14         
15         // bug 1: use error init.
16         int[][] D = new int[lenA + 1][lenB + 1];
17         
18         int max = 0;
19         
20         // BUG 2: should use <= instead of <
21         for (int i = 0; i <= lenA; i++) {
22             for (int j = 0; j <= lenB; j++) {
23                 if (i == 0 || j == 0) {
24                     D[i][j] = 0;
25                 } else {
26                     if (A.charAt(i - 1) == B.charAt(j - 1)) {
27                         D[i][j] = D[i - 1][j - 1] + 1;
28                     } else {
29                         D[i][j] = 0;
30                     }
31                 }
32                 
33                 max = Math.max(max, D[i][j]);
34             }
35         }
36         
37         return max;
38     }
39 }