718. Maximum Length of Repeated Subarray 最长重复子数组
【LeetCode】718. Maximum Length of Repeated Subarray 解题报告(Python)
标签(空格分隔): LeetCode
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/
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 <= len(A), len(B) <= 1000
- 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
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
参考资料: http://www.cnblogs.com/grandyang/p/7801533.html
日期
2018 年 9 月 11 日 ———— 天好阴啊