189. 크레인

1) 어떤 전략(알고리즘)으로 해결?

2) 코딩 설명

<내 풀이>

from collections import deque


def solution(board, moves):
    
    board = deque(board)
    boardlen = len(board)
    basket = deque()
    poppeddoll = 0

    for i in range(len(moves)) : 

        if board[boardlen-1][moves[i]-1]==0 : 
            #print("  moc", moves[i]-1)
            continue
        else :
            for j in range(boardlen) :
                if board[j][moves[i]-1]!=0:
                    basket.append(board[j][moves[i]-1])
                    #print("doll : ", board[j][moves[i]-1])
                    board[j][moves[i]-1]=0
                    if len(basket)>1 :
                        if basket[-1]==basket[-2] :
                            
                            basket.pop()
                            basket.pop()
                            poppeddoll+=2
                    break


    return poppeddoll

print(solution([[0,0,0,0,0],[0,0,1,0,3],[0,2,5,0,1],[4,2,4,4,2],[3,5,1,3,1]], 	[1,5,3,5,1,2,1,4]))

<내 틀렸던 풀이, 문제점>

from collections import deque

def solution(board, moves):
    answer = 0
    basket=deque()
    board = deque(board)
    
    for i in range(len(moves)) :
        
        print(answer, list(basket), moves[i]-1)    
        if board[moves[i]-1] : 
            doll = board[moves[i]-1].pop()
            if basket and basket[-1]==doll :
                basket.pop()
                answer+=2
            else : basket.append(doll)
    
    return answer

solution([[0,0,0,0,0],[0,0,1,0,3],[0,2,5,0,1],[4,2,4,4,2],[3,5,1,3,1]], [1,5,3,5,1,2,1,4])

• 배열을 잘못 생각함 ㅠㅠ

from collections import deque

def solution(board, moves):

    answer = 0
    basket=deque()
    board = deque(board)
    boardlen = len(board[0])
    
    for i in range(len(moves)) :
        if board[len(board)-1][moves[i]-1] == 0 : continue
        for j in range(boardlen) : 

            if board[j][moves[i]-1]==0 : continue
            else :
                if basket and basket[-1]==board[j][moves[i]-1] :
                    basket.pop()
                    answer+=2
                else : 
                    basket.append(board[j][moves[i]-1])
                    board[j][moves[i]-1]=0
                break

    return answer

solution([[0,0,0,0,0],[0,0,1,0,3],[0,2,5,0,1],[4,2,4,4,2],[3,5,1,3,1]], [1,5,3,5,1,2,1,4])

<반성 점 & 배운 점>

세게 배운 점 :

• 처음에 문제 이해하지 않고 생각없이 코딩부터 들이밀면 안된다.
• 설계, 문제이해 충분히 한 후 이를 구현하도록 하자
• 지금 코드가 이해안되고 답이 없으면 과감히 지워버리도록 하자