这是我参与8月更文挑战的第5天，活动详情查看：8月更文挑战
描述
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
Note:
1 <= nums.length <= 10^5
0 <= k <= nums.length
nums[i] is 0 or 1
解析
根据题意，就是判断 nums 中的 1 是否中间至少有 k 个 0 。这个题有点坑，题意有点没说明白，导致我两次提交错误，其实当 nums 为全 0 或者只有 1 个 1 的时候，这也算是 True ，卧槽无情。其他的就是正常的判断，遍历 nums ，找出两个为 1 的元素的索引差和 k 比较一下即可，不满足题意直接返回 False ，否则遍历结束之后直接返回 True 。
解答
class Solution(object):
def kLengthApart(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: bool
"""
if sum(nums)<=1:
return True
start = nums.index(1)
end = nums.index(1,start+1)
if end-start<k+1:
return False
index = end+1
while index<len(nums):
if nums[index]==1:
if index-end<k+1:
return False
end = index
index += 1
return True
运行结果
Runtime: 468 ms, faster than 95.71% of Python online submissions for Check If All 1's Are at Least Length K Places Away.
Memory Usage: 15.9 MB, less than 82.14% of Python online submissions for Check If All 1's Are at Least Length K Places Away.
解析
上面的代码其实比较乱，我又整理了一下，原理和上面一样，比较简洁易懂一点。
解答
class Solution(object):
def kLengthApart(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: bool
"""
count = k
for num in nums:
if num == 1:
if count < k:
return False
count = 0
else:
count += 1
return True
运行结果
Runtime: 468 ms, faster than 91.23% of Python online submissions for Check If All 1's Are at Least Length K Places Away.
Memory Usage: 15.8 MB, less than 78.95% of Python online submissions for Check If All 1's Are at Least Length K Places Away.
原题链接：leetcode.com/problems/ch…
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