79 最长公共子串 (lintcode)

 f[i][j]表示的是以第i个结尾和第j个结尾

class Solution {
public:
    /*
     * @param A: A string
     * @param B: A string
     * @return: the length of the longest common substring.
     */
    int longestCommonSubstring(string &A, string &B) {
        // write your code here
        int length1 = A.length();
        int length2 = B.length();
        vector<vector<int> > result(length1+1,vector<int>(length2+1));
        for(int i = 0;i <= length1;i++)
            result[i][0] = 0;
        for(int j = 0;j <= length2;j++)
            result[0][j] = 0;
        for(int i = 1;i <= length1;i++){
            for(int j = 1;j <= length2;j++){
                if(A[i-1] == B[j-1])
                    result[i][j] = result[i-1][j-1] + 1;
                else{
                    result[i][j] = 0;
                }
            }
        }
        int max = 0;
        for(int i = 1;i <= length1;i++){
            for(int j = 1;j <= length2;j++){
                if(result[i][j] > max)
                    max = result[i][j];
            }
        }
        return max;
    }
};

 

posted @ 2017-09-06 19:49  有梦就要去实现他  阅读(197)  评论(0编辑  收藏  举报