Find Numbers with Even Number of Digits
Description
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 <= 500
1 <= nums[i] <= 105
Solution(javascript)
/**
* @param {number[]} nums
* @return {number}
*/
var findNumbers = function(nums) {
let evenNums = []
for (let i = 0; i < nums.length; i++){
nums[i] = nums[i].toString();
if (nums[i].length % 2 === 0) evenNums.push(nums[i])
}
return evenNums.length;
};