import java.util.*; class Solution { public String solution(int[] numbers) { String answer = ""; String[] numString = new String[numbers.length]; for(int i=0; i

class Solution { public int solution(int n) { int answer = 0; int oneCount = 0; String biN = Integer.toBinaryString(n); for(int i=0; i

import java.util.*; class Solution { public int solution(int[][] baseball) { int answer = 0; List allCase = new ArrayList(); for(int i=123; i

import java.util.*; class Solution { public int solution(int[] people, int limit) { int answer = 0; int minPeople=0; int maxPeople=people.length-1; Arrays.sort(people); while(true){ if(minPeople==maxPeople){ answer++; break; } if(minPeople+1==maxPeople){ if(people[minPeople] + people[maxPeople]

import java.util.*; class Solution { public int solution(int[] nums) { int answer = 0; Set ponketmon = new HashSet(); for(int i=0; i

import java.util.*; public class Solution { public int solution(int n) { int ans = 1; while(n!=1){ if(n%2==0){ n=n/2; }else{ ans++; n=n-1; } } return ans; } } 도착점이 짝수이면 전에 도착한데까지에서 순간이동을 하면 되므로 소모되는 것이없다. 도착점이 홀수이면 (도착점-1) 에서 1만큼 소모하면 되므로 (도착점-1) 까지의 소모값에 1을 더해주면 된다.