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描述
Given an array of positive integers arr, find a pattern of length m that is repeated k or more times.
A pattern is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times consecutively without overlapping. A pattern is defined by its length and the number of repetitions.
Return true if there exists a pattern of length m that is repeated k or more times, otherwise return false.
Example 1:
Input: arr = [1,2,4,4,4,4], m = 1, k = 3
Output: true
Explanation: The pattern (4) of length 1 is repeated 4 consecutive times. Notice that pattern can be repeated k or more times but not less.
Example 2:
Input: arr = [1,2,1,2,1,1,1,3], m = 2, k = 2
Output: true
Explanation: The pattern (1,2) of length 2 is repeated 2 consecutive times. Another valid pattern (2,1) is also repeated 2 times.
Example 3:
Input: arr = [1,2,1,2,1,3], m = 2, k = 3
Output: false
Explanation: The pattern (1,2) is of length 2 but is repeated only 2 times. There is no pattern of length 2 that is repeated 3 or more times.
Example 4:
Input: arr = [1,2,3,1,2], m = 2, k = 2
Output: false
Explanation: Notice that the pattern (1,2) exists twice but not consecutively, so it doesn't count.
Example 5:
Input: arr = [2,2,2,2], m = 2, k = 3
Output: false
Explanation: The only pattern of length 2 is (2,2) however it's repeated only twice. Notice that we do not count overlapping repetitions.
Note:
2 <= arr.length <= 100
1 <= arr[i] <= 100
1 <= m <= 100
2 <= k <= 100
解析
根据题意，就是给出了一个整数数组 arr ，判断是否有长度为 m 的模式连着重复至少 k 次。这里看英文题目描述可能不是很清晰，但是看了例子就基本知道了，主要记住这 k 个模式不要断开，但也不能互相交叉覆盖。思路比较简单：
- 如果 m*k 大于 arr 的长度，说明无法满足题意，直接返回 False
- 初始化 idx 为 0 表示模式的起始索引，length 为 m*k 表示模式的长度
- 在 while 循环中，arr[idx:idx+length] 提取当前的子数组，然后判断其是否有满足题意的模式，如果有则直接返回 True ，如果没有则 idx 加一，判断以下一个索引为开头的子数组是否有满足题意的模式
- 当 idx+length>len(arr) 则遍历结束，返回 False
解答
class Solution(object):
def containsPattern(self, arr, m, k):
"""
:type arr: List[int]
:type m: int
:type k: int
:rtype: bool
"""
if len(arr) < m*k:
return False
idx = 0
length = m*k
while idx+length<=len(arr):
subarr = arr[idx:idx+length]
if subarr[:m] * k == subarr:
return True
idx += 1
return False
运行结果
Runtime: 20 ms, faster than 84.21% of Python online submissions for Detect Pattern of Length M Repeated K or More Times.
Memory Usage: 13.5 MB, less than 38.60% of Python online submissions for Detect Pattern of Length M Repeated K or More Times.
原题链接：leetcode.com/problems/de…
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