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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.
What‘s the definition of Longest Common Subsequence?
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:
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 if(A == null || B == null){ 8 return 0; 9 } 10 int m = A.length(); 11 int n = B.length(); 12 int [][] dp = new int[m+1][n+1]; 13 for(int i = 1; i<=m; i++){ 14 for(int j = 1; j<=n; j++){ 15 //两个末尾char match, 数目就是dp[i-1][j-1]+1 16 if(A.charAt(i-1) == B.charAt(j-1)){ 17 dp[i][j] = dp[i-1][j-1]+1; 18 }else{ 19 dp[i][j] = Math.max(dp[i][j-1],dp[i-1][j]); 20 } 21 } 22 } 23 return dp[m][n]; 24 } 25 }
LintCode Longest Common Subsequence
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5349730.html
