Codewars note: 奇偶判定

判断 奇数 偶数：

Solution:

 1

def even_or_odd(number):
    return 'Even' if number % 2 == 0 else 'Odd'

2

def even_or_odd(number):
    return ['Even', 'Odd'][number % 2]

标签：even,奇偶,return,odd,number,Codewars,note,Odd,Even
来源： https://www.cnblogs.com/a001ai-es/p/16462770.html