题目地址：https://leetcode.com/problems/maximum-length-of-repeated-subarray/description/
题目描述：
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)<=1000；
2、 0<=A[i],B[i]<100；
题目大意
求最长重复子数组。那么如果我们将数组换成字符串，实际这道题就是求Longest Common Substring的问题了。
解题方法
这个题显然是DP。一定注意，必须连续才行！那么dp数组中每个不为0的位置，一定是两者相等的地方。
比如，对于这两个数组[1,2,2]和[3,1,2]，我们的dp数组为：
3 1 2
1 0 1 0
2 0 0 2
2 0 0 1
所以递推关系为，dp[i][j] = dp[i-1][j-1]，当A[i]== B[j]。如果不等的话，dp[i][j]为0.
刚开始理解成了最长子序列Longest Common Subsequence问题了。耽误了不少时间……
代码如下：
class Solution:
def findLength(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: int
"""
m, n = len(A), len(B)
dp = [[0 for j in range(n + 1)] for i in range(m + 1)]
max_len = 0
for i in range(m + 1):
for j in range(n + 1):
if i == 0 or j == 0:
dp[i][j] = 0
elif A[i - 1] == B[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
max_len = max(max_len, dp[i][j])
return max_len
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换一种方式写，可能更好理解吧，毕竟少了一行和一列空的0.
class Solution:
def findLength(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: int
"""
m, n = len(A), len(B)
dp = [[0 for j in range(n)] for i in range(m)]
max_len = 0
for i in range(m):
for j in range(n):
if A[i] == B[j]:
if i == 0 or j == 0:
dp[i][j] = 1
else:
dp[i][j] = dp[i - 1][j - 1] + 1
max_len = max(max_len, dp[i][j])
return max_len
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