[BOJ/C++]28279번_ 덱 2(Deque)

https://www.acmicpc.net/problem/28279

##문제 풀기 전 내가 알고 있었어야 할 개념:

＊Deque의 [Rear]에서 Push(), Pop() 하는 메소드 => 'back'

• Push() : Deque객체.push_back
• Pop() : Deque객체.pop_back

 

#include <iostream>
#include <deque>
 
using namespace std;
 
deque <int> DQ;
 
int main() {
 
    ios::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
 
    int N,menu;
    //명령의 수 입력
    cin >> N;
 
    while (N--) {
        cin >> menu;
 
        if (menu == 1) {//front로 push
            int input;
            cin >> input;
            DQ.push_front(input);
        }
        else if (menu == 2) {//rear로 push
            int input;
            cin >> input;
            DQ.push_back(input);
        }
        else if (menu == 3) {//front값 pop
            if (DQ.empty()) cout << -1 << '\n';
            else {
                cout << DQ.front() << '\n';
                DQ.pop_front();
            }
        }
        else if (menu == 4) {//rear값 pop
            if (DQ.empty()) cout << -1 << '\n';
            else {
                cout << DQ.back() << '\n';
                DQ.pop_back();
            }
        }
        else if (menu == 5) {//size() 출력
            cout << DQ.size()<<'\n';
        }
        else if (menu == 6) {//empty 출력
            if (DQ.empty()) cout << 1 << '\n';
            else cout << 0 << '\n';
        }
        else if (menu == 7) {//front peek
            if (DQ.empty()) cout << -1 << '\n';
            else {
                cout << DQ.front() << '\n';
            }
        }
        else if (menu == 8) {//rear로 peek
            if (DQ.empty()) cout << -1 << '\n';
            else {
                cout << DQ.back() << '\n';
            }
        }
    }
 
    return 0;
}