python活力练习Day28

题目描述： 最长重复子数组

给两个整数数组 A 和 B ，返回两个数组中公共的、长度最长的子数组的长度。

示例 1:

输入:
A: [1,0,0,1,1]
B: [1,0,0,0,1]
输出: 3
解释: 
长度最长的公共子数组是 [1,0,0] or [0,0,1]。

说明:

1. 1 <= len(A), len(B) <= 1000
2. 0 <= A[i], B[i] < 100

引申概念：

最长公共子序列：

最长公共子串：

 python代码：

最长公共子串：

def findLength( A, B):
    n = len(A)
    m = len(B)
    dis = [[0 for i in range(n + 1)] for j in range(m + 1)]
    for i in range(m):
        for j in range(n):
            if A[i] == B[j]:
                dis[i + 1][j + 1] = dis[i][j] + 1
    max = 0
    for i in range(n+1):
        for j in range(n+1):
            if dis[i][j] > max:
                max = dis[i][j]

    return max

if __name__ == "__main__":

    A = [1, 0, 0, 1, 1]
    B = [1, 0, 0, 0, 1]
    print(findLength(A,B))

输出结果：3   此时最长公共子串为[1,0,0] 或者【0,0,1】

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

最长公共子序列：

def findLength( A, B):
    n = len(A)
    m = len(B)
    dis = [[0 for i in range(n + 1)] for j in range(m + 1)]
    for i in range(m):
        for j in range(n):
            if A[i] == B[j]:
                dis[i + 1][j + 1] = dis[i][j] + 1
            else:
                dis[i + 1][j + 1] = max(dis[i+1][j],dis[i][j+1])

    return dis[-1][-1]

if __name__ == "__main__":

    A = [1, 0, 0, 1, 1]
    B = [1, 0, 0, 0, 1]
    print(findLength(A,B))

输出结果：4   此时最长公共子序列为[1,0,0,1] 

		1	0	0	0	1
	0	0	0	0	0	0
1	0	1	1	1	1	1
0	0	1	2	2	2	2
0	0	1	2	3	3	3
1	0	1	2	3	3	4
1	0	1	2	3	3	4

定义来源：https://blog.csdn.net/ggdhs/article/details/90713154

题目来源：https://leetcode-cn.com/problems/maximum-length-of-repeated-subarray/