[프로그래머스 행렬 테두리 회전하기

문제

https://school.programmers.co.kr/learn/courses/30/lessons/77485

풀이

행렬 테두리의 값을 deque에 넣고 회전한 다음,
테두리에 재할당하도록 했다.

행렬 가로 세로 길이를 구하고, 순회하면서 테두리 값을 배열에 넣고
그 배열을 deque화 한다.

deque를 회전하고 다시 재할당 한다.

from collections import deque


def solution(rows, columns, queries):
    answer = []
    board = [[i * columns + j + 1 for j in range(columns)] for i in range(rows)]

    def selectTarget(querie):
        y1, x1, y2, x2 = [i - 1 for i in querie]
        target = []
        cur_y, cur_x = y1, x1
        size_y = y2 - y1 + 1
        size_x = x2 - x1 + 1

        for i in range(size_x):
            target.append(board[cur_y][cur_x])
            cur_x += 1

        cur_y, cur_x = cur_y + 1, cur_x - 1

        for i in range(size_y - 1):
            target.append(board[cur_y][cur_x])
            cur_y += 1

        cur_y, cur_x = cur_y - 1, cur_x - 1

        for i in range(size_x - 1):
            target.append(board[cur_y][cur_x])
            cur_x -= 1

        cur_y, cur_x = cur_y - 1, cur_x + 1

        for i in range(size_y - 2):
            target.append(board[cur_y][cur_x])
            cur_y -= 1

        answer.append(min(target))

        return target

    def rotate(dq, querie):
        dq.rotate(1)
        y1, x1, y2, x2 = [i - 1 for i in querie]
        cur_y, cur_x = y1, x1
        size_y = y2 - y1 + 1
        size_x = x2 - x1 + 1

        index = 0
        for i in range(size_x):
            board[cur_y][cur_x] = dq[index]
            cur_x += 1
            index += 1

        cur_y, cur_x = cur_y + 1, cur_x - 1

        for i in range(size_y - 1):
            board[cur_y][cur_x] = dq[index]
            cur_y += 1
            index += 1

        cur_y, cur_x = cur_y - 1, cur_x - 1

        for i in range(size_x - 1):
            board[cur_y][cur_x] = dq[index]
            cur_x -= 1
            index += 1

        cur_y, cur_x = cur_y - 1, cur_x + 1

        for i in range(size_y - 2):
            board[cur_y][cur_x] = dq[index]
            cur_y -= 1
            index += 1

    for querie in queries:
        dq = deque(selectTarget(querie))
        rotate(dq, querie)

    return answer

다른 사람의 풀이

stack 을 활용한 풀이.
테두리 값을 배열에 넣고 배열 길이가 2개 이상일 때만 회전이 가능하니
맨뒤에서 두번째 값(이전값)으로 재할당 한다.

def solution(rows, columns, queries):
    answer = []

    board = [[i+(j)*columns for i in range(1,columns+1)] for j in range(rows)]

    for a,b,c,d in queries:
        stack = []
        r1, c1, r2, c2 = a-1, b-1, c-1, d-1


        for i in range(c1, c2+1):

            stack.append(board[r1][i])
            if len(stack) == 1:
                continue
            else:
                board[r1][i] = stack[-2]


        for j in range(r1+1, r2+1):
            stack.append(board[j][i])
            board[j][i] = stack[-2]

        for k in range(c2-1, c1-1, -1):
            stack.append(board[j][k])
            board[j][k] = stack[-2]

        for l in range(r2-1, r1-1, -1):
            stack.append(board[l][k])
            board[l][k] = stack[-2]

        answer.append(min(stack))


    return answer