[leetcode] 1437. Check If All 1‘s Are at Least Length K Places Away

Description

Given an array nums of 0s and 1s and an integer k, return True if all 1’s are at least k places away from each other, otherwise return False.

Example 1:

Input: nums = [1,0,0,0,1,0,0,1], k = 2 Output: true Explanation: Each of the 1s are at least 2 places away from each other.

Example 2:

Input: nums = [1,0,0,1,0,1], k = 2 Output: false Explanation: The second 1 and third 1 are only one apart from each other.

Example 3:

Input: nums = [1,1,1,1,1], k = 0 Output: true

Example 4:

Input: nums = [0,1,0,1], k = 1 Output: true

Constraints:

1 <= nums.length <= 10^50 <= k <= nums.lengthnums[i] is 0 or 1

分析

题目的意思是：给你一个包含0，1的数组，判断1与1之间的间隔数是否都至少大于K，这道题思路很直接，就遍历判断就好了，我的时间复杂度O(n)，应该没有比这个更优的解法了，不太清楚为啥是middle的题目。

代码

class Solution: def kLengthApart(self, nums: List[int], k: int) -> bool: s=-1 n=len(nums) for i in range(n): if(nums[i]==1 and s==-1): s=i elif(nums[i]==1): interval=i-s-1 if(interval<k): return False s=i return True