---
tags: LeetCode
---

# 1295. Find Numbers with Even Number of Digits

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

1. 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.

2. 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

---

輸入範本如下

public class Solution {
    public int FindNumbers(int[] nums)
    {

    }
}

自行限制 , 不准使用 LinQ

public int FindNumbers(int[] nums)
{
     return nums.Count(num =>
     {
          bool isEven = true;
          while (num > 0)
          {
               num /= 10;
               isEven = !isEven;
          }
          return isEven;
     });

public int FindNumbers(int[] nums) => nums.Count(num => 
                                      num.ToString().Length % 2 == 0);

### 直覺想法
1. 第一個想法是我要怎麼知道一個數字的位數是多少!?
因為數字是十進位的 , 所以一個數字一直除以 10 可以除幾次 ,應該就代表它的位數了吧.

public int FindNumbers(int[] nums)
{
    int count = 0;
    foreach (var i in nums)
    {
        if (IsEvenDigit(i))
        {
            count++;
        }
    }
    return count;

    bool IsEvenDigit(int num)
    {
        bool isEven = true;
        while (num > 0)
        {
            num /= 10;
            isEven = !isEven;
        }
        return isEven;
    }
}

2. 後來的想法是將數字轉成字串 , 自然也就可以知道它的位數是多少了

public int FindNumbers(int[] nums)
{
     int count = 0;
     foreach (var i in nums)
     {
          if (IsEvenDigit(i))
          {
               count++;
          }
     }
     return count;

     bool IsEvenDigit(int num) => num.ToString().Length % 2 == 0;
}

### Thank you!

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