![[알고리즘문제풀기]짝수 홀수 개수](https://image.inblog.dev?url=https%3A%2F%2Finblog.ai%2Fapi%2Fog%3Ftitle%3D%255B%25EC%2595%258C%25EA%25B3%25A0%25EB%25A6%25AC%25EC%25A6%2598%25EB%25AC%25B8%25EC%25A0%259C%25ED%2592%2580%25EA%25B8%25B0%255D%25EC%25A7%259D%25EC%2588%2598%2520%25ED%2599%2580%25EC%2588%2598%2520%25EA%25B0%259C%25EC%2588%2598%26logoUrl%3Dhttps%253A%252F%252Finblog.ai%252Finblog_logo.png%26blogTitle%3D%25EC%2586%25A1%25EC%258A%25B9%25ED%2598%2584%25EC%259D%2598%2520%25EB%25B8%2594%25EB%25A1%259C%25EA%25B7%25B8&w=2048&q=75)
문제
나의 해답
class Solution {
public int[] solution(int[] num_list) {
int odd= 0;
int even= 0;
for(int i =0; i<num_list.length; i++){
if(!(num_list[i]%2==0)){
odd++;
}else{
even++;
}
}
int[] answer = {even,odd};
return answer;
}
}
해설
- 홀수를 카운트할 odd, 짝수를 카운트할 even 변수 생성
- 매개변수 num_list를 받아 num_list의 길이만큼 반복
- num_list[i]를 2로 나누어 나머지 값이 있으면 odd를 ++, 없으면 even을 ++
- odd와 even을 담은 정수 배열을 생성
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