프로그래머스 lv1 크레인인형뽑기게임, 키패드누르기

1. 개요

• https://school.programmers.co.kr/learn/courses/30/lessons/64061
• https://school.programmers.co.kr/learn/courses/30/lessons/67256

2. 크레인인형뽑기게임 풀이

def solution(board, moves):
    # board를 stack형태로 변경한다.
    board = [list(reversed(row)) for row in zip(*board)]
    stack = []
    answer = 0
    
    for move in [m-1 for m in moves]:
        while board[move] and board[move][-1] == 0:
            board[move].pop()

        if not board[move]:
            continue
        
        new = board[move].pop()
        
        if stack and stack[-1] == new:
            stack.pop()
            answer += 2
        else:
            stack.append(new)

    return answer

3. 키패드누르기 풀이

# 00 01 02    
# 10 11 12
# 20 21 22
# 30 31 32
def solution(numbers, hand):
    R, C = [3,0,0,0,1,1,1,2,2,2], [1,0,1,2,0,1,2,0,1,2]
    lr, lc, rr, rc = 3, 0, 3, 2
    left, right = {1, 4, 7}, {3, 6, 9}
    answer = []
    
    for n in numbers:
        if n in left:
            lr, lc = R[n], C[n]
            answer.append("L")
        elif n in right:
            rr, rc = R[n], C[n]
            answer.append("R")
        elif abs(lr-R[n])+abs(lc-C[n]) > abs(rr-R[n])+abs(rc-C[n]):
            rr, rc = R[n], C[n]
            answer.append("R")
        elif abs(lr-R[n])+abs(lc-C[n]) < abs(rr-R[n])+abs(rc-C[n]):
            lr, lc = R[n], C[n]
            answer.append("L")
        elif hand == "right":
            rr, rc = R[n], C[n]
            answer.append("R")
        else:
            lr, lc = R[n], C[n]
            answer.append("L")
    return "".join(answer)