[프로그래머스][Level1] 크레인 인형뽑기 게임 c++

https://programmers.co.kr/learn/courses/30/lessons/64061 

프로그래머스

 

풀이과정

moves횟수만큼 반복문을 돌려주고 moves가 해당하는 열에 0이 아닌 것이 있으면  
인형이 있다는 뜻이므로 스택에 push를 해준다. 만약 stack이 비어있지 않고  
top이 넣는 인형과 같으면 두 개씩 터지는 것이므로 2를 답에 더해준다.

여기서 가장 중요한 부분은 break인데  
0이 아닌 것을 찾았으면 한 번 실행만 하고 다음 move로 넘어가야 된다.

 

풀이1

#include <string>
#include <vector>
#include <stack>

using namespace std;

int solution(vector<vector<int>> board, vector<int> moves) {
    int answer = 0;

    stack<int> s;
    for(int i = 0; i < moves.size(); i++){
        for(int j = 0; j < board.size(); j++){
            if(board[j][moves[i] - 1] != 0) {
                if(!s.empty() && s.top() == board[j][moves[i] - 1]){
                    s.pop();
                    answer += 2;
                }
                else s.push(board[j][moves[i] - 1]);
                board[j][moves[i] - 1] = 0;
                break;
            }
        }
    }
    return answer;
}

 

풀이2

#include <string>
#include <vector>
#define STACK_SIZE 1001

using namespace std;
// 스택 구현
int top = -1;
int st[STACK_SIZE];

bool IsEmpty(){
    if(top < 0) return true;
    else return false;
}
void push(int n){
    st[++top] = n;
}
void pop(){
    if(!IsEmpty()) st[top--];
}
int peek(){
    if(!IsEmpty()) return st[top];
}

int solution(vector<vector<int>> board, vector<int> moves) {
    int answer = 0;

    for(int i = 0; i < moves.size(); i++){
        for(int j = 0; j < board.size(); j++){
            if(board[j][moves[i] - 1] != 0) {
                if(!IsEmpty() && peek() == board[j][moves[i] - 1]){
                    pop();
                    answer += 2;
                }
                else push(board[j][moves[i] - 1]);
                board[j][moves[i] - 1] = 0;
                break;
            }
        }
    }
    return answer;
}