[Java] LeetCode 1295 : Find Numbers with Even Number of Digits

LeetCode # 1295

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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

 

 

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Solution #1 (1ms)

class Solution {
    public int findNumbers(int[] nums) {
        int count = 0;
        for (int n : nums) {
            count = String.valueOf(n).length() % 2 == 0 ? count + 1 : count;
        }
        return count;
    }
}

 

Result #1

 

💡 간단하게 if 조건문과 삼항 연산자를 이용하여 풀었다. 

 

 

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Solution #2

class Solution {
    public int findNumbers(int[] nums) {
        int c = 0;
        for (int n : nums) {
            c = String.valueOf(n).length() % 2 == 0 ? c + 1 : c;
        }
        System.gc();
        return c;
    }
}

 

Result #2

 

💡 System.gc() 를 추가했더니 시간은 1ms 증가하고, 메모리는 감소하였다.

 

 

 

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More Algorithm!

 

👇👇

 

github.com/ggujangi/ggu.leet-code

ggujangi/ggu.leet-code

 

 

 

 

출처 : leetCode