[LeetCode] Find Numbers with Even Number of Digits

🔍  문제

Given an array nums of integers, return how many of them contain an even number of digits.

nums = [12,345,2,6,7896]
Output: 2
Explanation: 
12 contains 2 digits (even number of digits). 
345 contains 3 digits (odd number of digits). 
2 contains 1 digit (odd number of digits). 
6 contains 1 digit (odd number of digits). 
7896 contains 4 digits (even number of digits). 
Therefore only 12 and 7896 contain an even number of digits.

 

🖥  실행

1) Answer

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        cnt = 0
        evenList = []
        
        for i in nums:
            while i // 10 > 0:
                i = i //10
                cnt += 1
                
            cnt += 1    
            if cnt % 2 == 0:
                evenList.append(cnt)
            cnt = 0
            
        return len(evenList)

 

2) Better Solution

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
		return len([x for x in nums if len(str(x)) % 2 == 0])