lintcode-medium-Longest Common Subsequence

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

Your code should return the length of LCS.

Example

For

"ABCD"

and

"EDCA"

, the LCS is

"A"

(or

"D"

,

"C"

), return

1

.

For

"ABCD"

and

"EACB"

, the LCS is

"AC"

, return

2

.

动态规划：

用一个二维int数组表示A中前i个字符和B中前j个字符的LCS

状态转移：

当A的第i - 1个字符和B的第j - 1个字符相等时，dp[i][j] = dp[i - 1][j - 1] + 1

不相等时，选取dp[i - 1][j]和dp[i][j - 1]中较大的作为dp[i][j]的值

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) {
// write your code here

if(A == null || A.length() == 0 || B == null || B.length() == 0)
return 0;

int m = A.length();
int n = B.length();
int[][] dp = new int[m + 1][n + 1];

dp[0][0] = 0;
for(int i = 1; i <= m; i++)
dp[i][0] = 0;

for(int i = 1; i <= n; i++)
dp[0][i] = 0;

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] = dp[i - 1][j - 1] + 1;
else
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}

return dp[m]
;
}
}