Longest Common Subsequence
原题链接：http://lintcode.com/zh-cn/problem/longest-common-subsequence/
Given two strings, find the longest comment subsequence (LCS).
Your code should return the length of LCS.
样例
For "ABCD" and "EDCA", the LCS is "A" (or D or C), return 1
For "ABCD" and "EACB", the LCS is "AC", return 2
说明
What's the definition of Longest Common Subsequence?
* The longest common subsequence (LCS) problem is to find the longest subsequence common to all sequences in a set of sequences (often just two). (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.
* https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
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SOLUTION 1:
DP.
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， D[i - 1][j - 1] + 1
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 if (A == null || B == null) { 9 return 0; 10 } 11 12 int lenA = A.length(); 13 int lenB = B.length(); 14 int[][] D = new int[lenA + 1][lenB + 1]; 15 16 for (int i = 0; i <= lenA; i++) { 17 for (int j = 0; j <= lenB; j++) { 18 if (i == 0 || j == 0) { 19 D[i][j] = 0; 20 } else { 21 if (A.charAt(i - 1) == B.charAt(j - 1)) { 22 D[i][j] = D[i - 1][j - 1] + 1; 23 } else { 24 D[i][j] = Math.max(D[i - 1][j], D[i][j - 1]); 25 } 26 } 27 } 28 } 29 30 return D[lenA][lenB]; 31 } 32 }