[알고리즘문제풀기] n개 간격의 원소들

문제

내가 작성한 정답

1. Array이용

class Solution {
    public int[] solution(int[] num_list, int n) {
        int length = num_list.length%n==0?num_list.length/n:(num_list.length/n)+1;
        int[] answer = new int[length];
        int index = 0;
        for(int i =0; i<num_list.length; i+=n){
            answer[index++] = num_list[i];
        }
        return answer;
    }
}

2. ArrayList이용

import java.util.*;

class Solution {
    public int[] solution(int[] num_list, int n) {
        List<Integer> answer = new ArrayList<>();
        for(int i=0;i<num_list.length; i+=n){
            answer.add(num_list[i]);
        }
        int[] answer1 = new int[answer.size()];
        for(int i=0;i<answer.size(); i++){
            answer1[i] = answer.get(i);
        }
        return answer1;
    }
}

import java.util.*;

class Solution {
    public int[] solution(int[] num_list, int n) {
        List<Integer> answer = new ArrayList<>();
        for(int i=0;i<num_list.length; i+=n){
            answer.add(num_list[i]);
        }
        return answer.stream().mapToInt(Integer::intValue).toArray();
    }
}

3. StreamAPI이용

import java.util.stream.IntStream;

class Solution {
    public int[] solution(int[] num_list, int n) {

        return IntStream.range(0,num_list.length)
                .filter(i->i%n==0)
                .map(i->num_list[i])
                .toArray();
    }
}

다른 사람들의 정답

class Solution {
    public int[] solution(int[] num_list, int n) {
        Double length = Math.ceil(num_list.length/(n*1.0));
        int[] answer = new int[length.intValue()];

        for(int idx=0; idx<length; idx++) {
            answer[idx] = num_list[idx*n];
        }

        return answer;
    }
}

  double number1 = 5.3;
  double number2 = -2.7;

  // Math.ceil 사용
  double result1 = Math.ceil(number1); // 6.0
  double result2 = Math.ceil(number2); // -2.0