002 Find Numbers with Even Number of Digits

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] <= 10^5

Solution

Code

C#

public class Solution {
    public int FindNumbers(int[] nums) {
        int result = 0; //number for return
        int digits = 0; //number for counting
        int temp =0;
        
        for(int i=0; i<nums.Length; i++){
            temp= nums[i];
            //Get the digits of nums[i]
            for(int j =0; j<6; j++){
              if (temp > 0){
                temp = temp/10;
                ++digits;    //Get the number of digits
                }
              }
              
            //Judge whether it is even
            if (digits % 2 == 0){
                result++;
            }
            digits=0; // reset digits;
        }
    return result;
    }
}