Maximum Length of Repeated Subarray

Given two integer arraysAandB, 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. 1

   <

   = len(A), len(B)

   <

   = 1000

2. 0

   <

   = A[i], B[i]

   <

   100

分析

初始化为0，如果不等就更新为0. 最后结果取某段max

class Solution:
    def findLength(self, A: List[int], B: List[int]) -> int:
        n,m = len(A),len(B)
        f = [[0]*(m+1) for _ in range(n+1)]
        #f[0][0] = 0
        res = 0
        for i in range(1,n+1):
            #f[i][0] = f[i-1][0] + ord(s1[i-1])
            for j in range(1,m+1):
                #f[0][j] = f[0][j-1] + ord(s2[j-1])
                if A[i-1] == B[j-1]:
                    f[i][j] = f[i-1][j-1]+1
                    res = max(f[i][j],res)

        return res