Given an array nums of integers, return how many of them contain an even number of digits.
Example 1:
Input: 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.
Example 2:
Input: nums = [555,901,482,1771] Output: 1 Explanation: Only 1771 contains an even number of digits.
Constraints:
1 <= nums.length <= 5001 <= nums[i] <= 10^5
Solution: Math
Time complexity: O(n * log(max(num)))
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int findNumbers(vector<int>& nums) { return count_if(begin(nums), end(nums), [](int num) { int d = 0; do { ++d; } while (num /= 10); return d % 2 == 0; }); } }; |
Python3
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# Author: Huahua class Solution: def findNumbers(self, nums: List[int]) -> int: return sum([len(str(x)) % 2 == 0 for x in nums]) |
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