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Lintcode: Longest Common Subsequence

时间:2015-03-09 09:19:31      阅读:280      评论:0      收藏:0      [点我收藏+]
Given two strings, find the longest comment subsequence (LCS).

Your code should return the length of LCS.

Example
For "ABCD" and "EDCA", the LCS is "A" (or D or C), return 1

For "ABCD" and "EACB", the LCS is "AC", return 2

题目里说了:

Clarification
What‘s the definition of Longest Common Subsequence?

*(Note that a subsequence is different from a substring, for the terms of the former need not be consecutive terms of the original sequence.) It is a classic computer science problem, the basis of file comparison programs such as diff, and has applications in bioinformatics.

1. D[i][j] 定义为s1, s2的前i,j个字符串的最长common subsequence.

2. D[i][j] 当char i == char j, 可以有三种选择,D[i - 1][j - 1] + 1,D[i ][j - 1], D[i - 1][j] ,取最大的

    当char i != char j, D[i ][j - 1], D[i - 1][j] 里取一个大的(因为最后一个不相同,所以有可能s1的最后一个字符会出现在s2的前部分里,反之亦然。

 1 public class Solution {
 2     /**
 3      * @param A, B: Two strings.
 4      * @return: The length of longest common subsequence of A and B.
 5      */
 6     public int longestCommonSubsequence(String A, String B) {
 7         // write your code here
 8         int[][] res = new int[A.length()+1][B.length()+1];
 9         for (int i=1; i<=A.length(); i++) {
10             for (int j=1; j<=B.length(); j++) {
11                 res[i][j] = Math.max(A.charAt(i-1)==B.charAt(j-1)? res[i-1][j-1]+1 : res[i-1][j-1], 
12                 Math.max(res[i-1][j], res[i][j-1]));
13             }
14         }
15         return res[A.length()][B.length()];
16     }
17 }

 

Lintcode: Longest Common Subsequence

原文:http://www.cnblogs.com/EdwardLiu/p/4322681.html

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