Lintcode Longest Common subsequence

The original title link is here: http://www.lintcode.com/en/problem/longest-common-subsequence/

Topic:

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

Your code should return the length of the 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" "EDCA" and, the LCS is "A" (or "D" , "C" ), return 1 .

"ABCD"for and "EACB" , the LCS "AC" is, return 2 .

Exercises

Dp. Reference http://www.geeksforgeeks.org/dynamic-programming-set-4-longest-common-subsequence/

DP[I][J] Represents the length of the str1 length of I and the length of the str2 LCS of J.

If Str1.charat (i-1) = = Str2.charat (j-1) tail character is the same, dp[i][j] = dp[i-1][j-1]+1.

If different dp[i][j] = Math.max (Dp[i-1][j], dp[i][j-1]).

Example: 1. "Aggtab" and "Gxtxayb". Last characters match for the strings. So length of the LCS can be written as:
L ("Aggtab", "gxtxayb") = 1 + L ("Aggtab", "Gxtxayb")

2. "Abcdgh" and "AEDFHR". Last characters does not match for the strings. So length of the 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 classSolution {2     /**3      * @paramA, b:two strings.4      * @return: The length of longest common subsequence of A and B.5      */6      Public intlongestcommonsubsequence (String A, String B) {7         if(A = =NULL|| B = =NULL){8             return0;9         }Ten         intm =a.length (); One         intn =b.length (); A         int[] DP =New int[M+1] [N+1]; -          for(inti = 1; i<=m; i++){ -              for(intj = 1; j<=n; J + +){ the                 //two end char match, number is 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]); -                 } +             } A         } at         returnDp[m][n]; -     } -}

Lintcode Longest Common subsequence