LeetCode 718 Maximum Length of Repeated Subarray

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本文介绍了一种使用动态规划方法寻找两个整数数组间最长公共子数组的算法，并提供了一个具体的Python实现示例。

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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 <= len(A), len(B) <= 1000
0 <= A[i], B[i] < 100

假设A[i]结尾的子数组与B[j]结尾的子数组最大公共串为sub(i,j)，且公共串最后一个字符是A[i],B[j].

if A[i] == B[j]:

sub(i,j) = sub(i-1,j-1)+1

否则：

列一下sub的矩阵：

0	0	1	0	0
0	1	0	2	0
1	0	0	0	3
0	0	0	0	0
0	0

class Solution(object):
    def findLength(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: int
        """
        sub = [[0 for _ in range(len(A))] for _ in range(len(A))]
        res = 0
        for i in range(len(A)):
            for j in range(len(B)):
                if A[i] == B[j]:
                    if i==0 or j ==0:
                        sub[i][j] = 1
                    else:
                        sub[i][j] = sub[i-1][j-1]+1
                    res = max(res,sub[i][j])
        return res