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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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/find-numbers-with-even-number-of-digits/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    int getCount(int n) {
        int cnt = 0;
        while (n) {
            n /= 10;
            ++cnt;
        }
        return cnt;
    }
public:
    int findNumbers(vector<int>& nums) {
        int cnt = 0;
        for (int n : nums) {
            if (getCount(n) % 2 == 0) cnt++;
        }
        return cnt;
    }
};