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

题目意思是给定两个整型数组A和B，返回两个数组的公共子串的最大长度。
样例输入如下：

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

题目意思很明确，最长公共子数组问题,LCS用动态规划做比较合适。
取dp[i][j]表示A 数组到i位置，B数组到j位置时最长公共子数组长度，
转移方程为：

     if(A[i-1]==B[j-1]) {
          dp[i][j] = dp[i-1][j-1]+1;
    }  else{
          dp[i][j] = 0;
    }

然后用max变量记录最大的dp[i][j]
代码如下:

class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        
        int lenA = A.size();
        int lenB = B.size();
        vector<vector<int>> dp(lenA+1,vector<int>(lenB+1,0)); // 初始化数组为0
        
        int max = 0; // 记录最大长度
        
        for (int i = 1;i <= lenA ;i ++) {
            for (int j = 1;j<=lenB;j ++) {
                if(A[i-1] ==  B[j-1]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                    if(dp[i][j] > max) max = dp[i][j];
                }else {
                    dp[i][j] = 0;
                }
            }
        }
        return max; 
    }
};