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;
}
};
