Longest Common Subsequence

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

state: dp[i][j] 表示前i个字符配上前j个字符的LCS长度(必须包括以i j结尾的字符)

function： dp[i][j] 有两种情况:

               1. charAt(i) = charAt(j) dp[i][j] = dp[i-1][j-1], dp[i][j-1], dp[i-1][j]中最大的一个

               2. charAt(i) != charAt(j) dp[i][j] =dp[i][j-1], dp[i-1][j]中最大的一个 (只有最后一个不同, 可能i字符和j-1相同 或者j和i-1相同)

initial: 二维数组所以dp[i][0] = 0 dp[0][j] = 0

return: dp[m][n]

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++) {
                if (A.charAt(i - 1) == B.charAt(j - 1)) {
                    dp[i][j] = Math.max(dp[i-1][j-1] + 1, Math.max(dp[i][j-1], dp[i-1][j]));
                } else {
                    dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j]);
                }
            }
        }
        return dp[m][n];
    }
}