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

Example 1:
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].
Note:
1 <= len(A), len(B) <= 1000
0 <= A[i], B[i] < 100
就是類似找最長公共子序列的題目，利用動態規劃演算法。找出對大的那個值，複雜度為o（n^2）。遞推式：

if (A[i-1] == B[j-1])
match[i][j] = match[i - 1][j - 1] + 1;

class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int match[1001][1001] = {0};
        int max = 0;
        for (int i = 0; i <= A.size(); i++) {
            for (int j = 0; j <= B.size(); j++) {
                if
 
 (i == 0 || j == 0) {
                    match[i][j] = 0;
                }
                else {
                    if (A[i-1] == B[j-1]) {
                        match[i][j] = match[i - 1][j - 1] + 1;
                        if (match[i][j] > max) {
                            max = match[i][j];
                        }
                    }
                }

            }
        }
        return
 
 max;
    }
};