lintcode-medium-Longest Common Substring

Given two strings, find the longest common substring.

Return the length of it.

 

Example

Given A = "ABCD", B = "CBCE", return 2.

Challenge

O(n x m) time and memory.

 

这道题和longest common subsequence差不多，也是动态规划，状态转移稍有区别：

一个二维int数组记录A前i个字符和B前j个字符的longest common subsequence

当A的第i - 1个字符和B的第j - 1个字符不相等时，dp[i][j]为之前的情况中数值最大的，即dp[i][j - 1]和dp[i - 1][j]中较大的

当两个字符相等时，必须从后往前数包括这两个字符的substring的长度是否大于dp[i][j - 1]和dp[i - 1][j]，大于的话更新一下dp[i][j]，否则和不相等时一样

public class Solution {
    /**
     * @param A, B: Two string.
     * @return: the length of the longest common substring.
     */
    public int longestCommonSubstring(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] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
                else{
                    int len = 0;
                    while(i - len - 1 >= 0  && j - 1 - len >= 0 && A.charAt(i - 1 - len) == B.charAt(j - 1 - len))
                        len++;
                    
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                    dp[i][j] = Math.max(dp[i][j], len);
                }
                
            }
        }
        
        return dp[m][n];
    }
}