Longest Common String

Given two strings, find the longest common substring.

Return the length of it.

 Notice

The characters in substring should occur continuously in original string. This is different with subsequence.

Example

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

Analyse:

1. We can use brute force to solve this problem. Start from A[0...lengthA], to B[0...lengthB], if there is a character of the same, check its following and update the result.

Runtime: 19ms

class Solution {
public:    
    /**
     * @param A, B: Two string.
     * @return: the length of the longest common substring.
     */
    int longestCommonSubstring(string &A, string &B) {
        // write your code here
        int lengthA = A.size(), lengthB = B.size();
        if (!lengthA || !lengthB) return 0;
        
        int result = 0;
        for (int i = 0; i < lengthA; i++) {
            for (int j = 0; j < lengthB; j++) {
                if (A[i] == B[j]) {
                    int count = 1;
                    while (i + count < lengthA && j + count < lengthB && A[i + count] == B[j + count]) {
                        count++;
                    }
                    result = max(result, count);
                }
            }
        }
        return result;
    }
};

2. We can use dynamic programming to solve this problem. Suppose dp[i][j] is the longest common substring from A[0...i-1] and B[0...j-1]. If A[i] == B[j], then dp[i + 1][j + 1] = dp[i][j] + 1; else, dp[i + 1][j + 1] = 0.

Runtime: 19ms

class Solution {
public:    
    /**
     * @param A, B: Two string.
     * @return: the length of the longest common substring.
     */
    int longestCommonSubstring(string &A, string &B) {
        // write your code here
        int lengthA = A.size(), lengthB = B.size();
        if (!lengthA || !lengthB) return 0;
        
        vector<vector<int> > dp(lengthA + 1, vector<int>(lengthB + 1, 0));
        int result = INT_MIN;
        for (int i = 0; i < lengthA; i++) {
            for (int j = 0; j < lengthB; j++) {
                if (A[i] == B[j]) {
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                    result = max(result, dp[i + 1][j + 1]);
                }
            }
        }
        return result;
    }
};