本文正在参加「Python主题月」，详情查看 活动链接
描述
Given an integer array arr, return true if there are three consecutive odd numbers in the array. Otherwise, return false.
Example 1:
Input: arr = [2,6,4,1]
Output: false
Explanation: There are no three consecutive odds.
Example 2:
Input: arr = [1,2,34,3,4,5,7,23,12]
Output: true
Explanation: [5,7,23] are three consecutive odds.
Note:
1 <= arr.length <= 1000
1 <= arr[i] <= 1000
解析
根据题意，就是判断是否 arr 中存在三个相连的奇数。直接遍历 arr ，用 count 记录出现的奇数个数，如果是奇数则加一，如果不是奇数则重新置为 0 ，如果 count 大于等于 3 的时候直接返回 True ，否则最后遍历结束直接返回 False 。
解答
class Solution(object):
def threeConsecutiveOdds(self, arr):
"""
:type arr: List[int]
:rtype: bool
"""
count = 0
for i,v in enumerate(arr):
if v%2:
count+=1
if count==3:
return True
else:
count = 0
return False
运行结果
Runtime: 32 ms, faster than 67.58% of Python online submissions for Three Consecutive Odds.
Memory Usage: 13.4 MB, less than 91.78% of Python online submissions for Three Consecutive Odds.
解析
上面这种思路比较中规中矩，一般人看完题目之后都能想到，下面这种虽然原理是一样的，但是解法比较新奇，因为奇数的特性就是对 2 取模得 1 ，而 3 个相连的奇数对 2 取模就要出现 3 个 1 ，将 arr 中的每个数都取模得到的结果变成字符串连接成字符串，然后只需要判断字符串 111 是否在其中即可。
解答
class Solution(object):
def threeConsecutiveOdds(self, arr):
"""
:type arr: List[int]
:rtype: bool
"""
return "111" in "".join([str(i%2) for i in arr])
运行结果
Runtime: 36 ms, faster than 36.54% of Python online submissions for Three Consecutive Odds.
Memory Usage: 13.4 MB, less than 67.31% of Python online submissions for Three Consecutive Odds.
解析
这种算法是用到了比较少见的按位与运算 & ，因为某数字与 1 进行按位与运算的时候，当该数字是奇数的时候计算结果为 1 ，当该数字是偶数的时候计算结果为 0 ，所以可以沿用上面的思路，先对 arr 中的每个元素都对 1 做按位与运算，然后将得到的结果拼接为字符串，只需要判断字符串 111 是否在其中出现即可。
解答
class Solution(object):
def threeConsecutiveOdds(self, arr):
"""
:type arr: List[int]
:rtype: bool
"""
return '111' in ''.join([str(i&1) for i in arr])
运行结果
Runtime: 36 ms, faster than 36.54% of Python online submissions for Three Consecutive Odds.
Memory Usage: 13.5 MB, less than 67.31% of Python online submissions for Three Consecutive Odds.
原题链接：leetcode.com/problems/th…
您的支持是我最大的动力
