LeetCode：Maximum Length of Repeated Subarray

题目：

Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.

Examples:

Input:
A: [1,2,3,2,1]
B: [3,2,1,4,7]
Output: 3
Explanation:
The repeated subarray with maximum length is [3, 2, 1].

思路：

　　用一个二维数组 C[i][j] 来表示以A[i-1] 和 B[j-1] 结尾的最长子串的长度。显然，C[0][0]=0。当A[i-1] = B[j-1]时，C[i][j] = C[i-1][j-1] + 1，否则为0.（例如：如果A[0] = B[0]，那么C[1][1] = 1；如果 A[0] ！= B[0]，那么C[1][1] = 0）
　　

代码：

class Solution {public:    int findLength(vector<int>& A, vector<int>& B) {        int len1 = A.size();        int len2 = B.size();        vector<vector<int>> C(len1+1, vector<int>(len2+1, 0));        int len = 0;        for (int i = 1; i <= len1; i++) {            for (int j = 1; j <= len2; j++) {                if (A[i-1] == B[j-1]) {                    C[i][j] = C[i-1][j-1] + 1;                }                len = len > C[i][j] ? len : C[i][j];            }        }        return len;    }};