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];
    }
}