> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/l4dong-tai-gui-hua/maximum-length-of-repeated-subarray.md).

# Maximum Length of Repeated Subarray

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. 1 

   <

   \= len(A), len(B)&#x20;

   <

   \= 1000
2. 0&#x20;

   <

   \= A\[i], B\[i]&#x20;

   <

   &#x20;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
```