[python | 프로그래머스] LV 2. [1차] 프렌즈4블록

💗 문제

프로그래머스

 

💗 코드 미리보기

def solution(m, n, board):
    answer = 0
    board = [list(i) for i in board]

    while True:
        checked = [[0 for _ in range(n)] for _ in range(m)]

        for i in range(m - 1):
            for j in range(n - 1):
                if board[i][j] != 0 and board[i][j] == board[i + 1][j] == board[i][j + 1] == board[i + 1][j + 1]:
                    checked[i][j] = 1
                    checked[i][j + 1] = 1
                    checked[i + 1][j] = 1
                    checked[i + 1][j + 1] = 1

        cnt = 0
        for i in range(m):
            for j in range(n):
                if checked[i][j] == 1:
                    cnt += 1
                    board[i][j] = 0

        for i in range(m):
            for j in range(n):
                if board[i][j] != 0:
                    for a in range(i+1, m):
                        if checked[a][j] == 1:
                            board[i][j], board[a][j] = board[a][j], board[i][j]

        if cnt == 0:
            break
        else:
            answer += cnt

    return answer

 

💗 코드 쪼개보기

2x2 형태로 4개가 붙어있어 제거해야하는 블록들을 checked를 통해 처리함

        checked = [[0 for _ in range(n)] for _ in range(m)]

        for i in range(m - 1):
            for j in range(n - 1):
                if board[i][j] != 0 and board[i][j] == board[i + 1][j] == board[i][j + 1] == board[i + 1][j + 1]:
                    checked[i][j] = 1
                    checked[i][j + 1] = 1
                    checked[i + 1][j] = 1
                    checked[i + 1][j + 1] = 1

checked가 0인 경우, 제거해야할 블록이 더이상 없는 것이므로 함수를 빠져나가도록!

        cnt = 0
        for i in range(m):
            for j in range(n):
                if checked[i][j] == 1:
                    cnt += 1
                    board[i][j] = 0


        if cnt == 0:
            break
        else:
            answer += cnt

블록을 아래로 내려주는 함수

        for i in range(m):
            for j in range(n):
                if board[i][j] != 0:
                    for a in range(i+1, m):
                        if checked[a][j] == 1:
                            board[i][j], board[a][j] = board[a][j], board[i][j]