LintCode Longest Common Subsequence

原题链接在这里：http://www.lintcode.com/en/problem/longest-common-subsequence/

题目：

Given two strings, find the longest common subsequence (LCS).

Your code should return the length of LCS.

Clarification

What's the definition of Longest Common Subsequence?

• https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
• http://baike.baidu.com/view/2020307.htm

Example

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

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

题解：

DP. 参考http://www.geeksforgeeks.org/dynamic-programming-set-4-longest-common-subsequence/

dp[i][j]表示长度为i的str1 和 长度为j的str2 LCS长度.

若是str1.charAt(i-1)  == str2.charAt(j-1) 尾字符相同, dp[i][j] = dp[i-1][j-1]+1.

若是不同dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]).

举例: 1. "AGGTAB" and "GXTXAYB". Last characters match for the strings. So length of LCS can be written as:
L("AGGTAB", "GXTXAYB") = 1 + L("AGGTAB", "GXTXAYB")

2. "ABCDGH" and "AEDFHR". Last characters do not match for the strings. So length of LCS can be written as:
L("ABCDGH","AEDFHR") = MAX ( L("ABCDG", "AEDFHR"), L("ABCDGH", "AEDFH")).

Time Complexity: O(m*n). Space: O(m*n).

AC Java:

 public class Solution {
     /**
      * @param A, B: Two strings.
      * @return: The length of longest common subsequence of A and B.
      */
     public int longestCommonSubsequence(String A, String B) {
         if(A == null || B == null){
             return 0;
         }
         int m = A.length();
         int n = B.length();
         int [][] dp = new int[m+1][n+1];
         for(int i = 1; i<=m; i++){
             for(int j = 1; j<=n; j++){
                 //两个末尾char match, 数目就是dp[i-1][j-1]+1
                 if(A.charAt(i-1) == B.charAt(j-1)){
                     dp[i][j] = dp[i-1][j-1]+1;
                 }else{
                     dp[i][j] = Math.max(dp[i][j-1],dp[i-1][j]);
                 }
             }
         }
         return dp[m][n];
     }
 }