[프로그래머스][Level2] 기능 개발 c++

https://programmers.co.kr/learn/courses/30/lessons/42586 

프로그래머스

 

풀이 과정

이 문제는 큐를 활용하는 문제이다.  
각 기능들이 며칠째에 배포가 가능한지 큐에 넣는다.

그리고 만약 현재값보다 나중값이 더 작다면 같이 배포할 수 있기 때문에  
반복문을 돌려 몇개까지 배포할 수 있는지 res를 카운트 해준다.

 

풀이1

#include <string>
#include <vector>
#include <queue>

using namespace std;

vector<int> solution(vector<int> progresses, vector<int> speeds) {
    vector<int> answer;

    queue<int> q;
    for(int i = 0; i < progresses.size(); i++){
        int cnt = 0;
        while(progresses[i] < 100){
            progresses[i] += speeds[i];
            cnt++;
        }
        q.push(cnt);
    }
    while(!q.empty()){
        int now = q.front();
        q.pop();

        int res = 1;
        while(!q.empty() && now >= q.front()){
            res++;
            q.pop();
        }
        answer.push_back(res);
    }
    return answer;
}

 

 

풀이2

#include <string>
#include <vector>
#include <iostream>

using namespace std;

vector<int> solution(vector<int> progresses, vector<int> speeds) {
    vector<int> answer;
    
    vector<int> v; // 걸리는 일수
    for(int i = 0; i < progresses.size(); i++){
        int day = 0;
        int num = 100 - progresses[i];

        if (num % speeds[i] != 0) day += 1;
        day += num / speeds[i]; 
        
        v.push_back(day); 
    }
    
    int cnt = 1;
    int num = v[0]; // 젤 마지막 값
    for(int i = 1; i < v.size(); i++){
        if (num < v[i]){
            answer.push_back(cnt);
            cnt = 1;
            num = v[i];
        } else {
            cnt++;
        }
    }
    answer.push_back(cnt);
    
    return answer;
}