[프로그래머스] - 실패율

10 May 2021

Reading time ~1 minute

풀이 - 다시풀어보자

#include <string>
#include <vector>
#include <algorithm>
using namespace std;

bool compare(const pair<int,double> &a, const pair<int,double> &b){
    if(a.second>b.second){
        return true;
    }else if(a.second==b.second){
        if(a.first<b.first){
            return true;
        }else{
            return false;
        }
    }else{
        return false;
    }
}

vector<int> solution(int N, vector<int> stages) {
    vector<int> answer;
    int count=stages.size();
    vector <pair<int,double>> loss; //(n번 스테이지, 실패율)쌍
    int j=0;
    
    sort(stages.begin(),stages.end());

    for(int i=1; i<=N; i++){ //index i번째 stage부터 차례로
        if(count==0){
            loss.push_back(make_pair(i,0));
            continue;
        }

        int clear=0;
        
        while(stages[j]==i && j<stages.size()){
            clear++;
            j++;
        }

        double rate = (double) clear/count;
        loss.push_back(make_pair(i,rate));
        count -= clear;
    }
    
    sort(loss.begin(), loss.end(), compare);
    
    for(int i=0; i<N; i++){
        answer.push_back(loss[i].first);
    }

    return answer;
}

내가 풀다 만 코드 - 조금만 더 생각하고 고치면 될거같은데..

 #include <string>
#include <vector>
#include <algorithm>
using namespace std;

bool compare(pair<int,double>a, pair<int,double>b){
    if(a.first>b.first){
        return true;
    }else if(a.first==b.first){
        if(a.second<b.second){
            return true;
        }else{
            return false;
        }
    }else{
        return false;
    }
}

vector<int> solution(int N, vector<int> stages) {
    vector<int> answer;
    int count=stages.size();
    int clear=0;
    vector <pair<int,double>> loss; //(n번 스테이지, 실패율)쌍
    for(int i=0; i<=N; i++){
         loss.push_back(make_pair(i+1,0));
    }
   
    for(int i=0; i<=N; i++){ //index i번째 stage부터 차례로
        for(int j=0; j<stages.size(); j++){ //stage 배열 개수만큼 반복문
            if(loss[i].first == stages[j]){
                loss[i].second += (double)(1/count);
                clear++;
            }
        }
        count -= clear;
    }
    
    sort(loss.begin(), loss.end(), compare);
    
    for(int i=0; i<N; i++){
        answer.push_back(loss[i].first);
    }

    return answer;
} 