【leetcode】1566. Detect Pattern of Length M Repeated K or More Times
题目如下:
Given an array of positive integers
arr
, find a pattern of lengthm
that is repeatedk
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 lengthm
that is repeatedk
or more times, otherwise returnfalse
.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.Constraints:
2 <= arr.length <= 100
1 <= arr[i] <= 100
1 <= m <= 100
2 <= k <= 100
解题思路:本题是要判断arr中是否存在这样的子数组,其长度是m*k,并且子数组可以分成长度为m的k个孙数组,满足k个数组的元素和顺序完全一致。假设arr[i]是这样子数组的第一个元素,只要判断 arr[i:i+m] * k == arr[i:i+m*k] 即可。
代码如下:
class Solution(object): def containsPattern(self, arr, m, k): """ :type arr: List[int] :type m: int :type k: int :rtype: bool """ for i in range(len(arr)-m): if i+m*k <= len(arr) and arr[i:i+m] * k == arr[i:i+m*k]: return True return False