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].

 

 

Since a common subarray of A and B must start at some A[i] and B[j], let dp[i][j] be the length of longest common prefix of A[i:] and B[j:]. Whenever A[i] == B[j], we know dp[i][j] = dp[i+1][j+1] + 1. Also, the answer is           max(dp[i][j]) over all i, j.

class Solution {
    public int findLength(int[] A, int[] B) {
        if(A == null||B == null) return 0;
        int m = A.length;
        int n = B.length;
        int max = 0;
        //dp[i][j] is the length of longest common subarray ending with nums[i] and nums[j]
        int[][] dp = new int[m + 1][n + 1];
        for(int i = 0;i <= m;i++){
            for(int j = 0;j <= n;j++){
                if(i == 0 || j == 0){
                    dp[i][j] = 0;
                }
                else{
                    if(A[i - 1] == B[j - 1]){
                        dp[i][j] = 1 + dp[i - 1][j - 1];
                        max = Math.max(max,dp[i][j]);
                    }
                }
            }
        }
        return max;
    }
}