프로그래머스 - 행렬 테두리 회전하기 (L2)

 1dx = [1, 0, -1, 0]
 2dy = [0, 1, 0, -1]
 3
 4def solution(rows, columns, queries):
 5    table = [[(j*columns)+i+1 for i in range(columns)] for j in range(rows)]
 6    answer = []
 7
 8    for y1, x1, y2, x2 in queries:
 9        min_num = 10000
10        x1 -= 1
11        y1 -= 1
12        x2 -= 1
13        y2 -= 1
14        direction = 0
15        curX, curY = x1, y1
16        postNum = table[y1][x1]
17        while True:
18            curX = curX+dx[direction]
19            curY = curY+dy[direction]
20
21            temp = table[curY][curX]
22            table[curY][curX] = postNum
23            postNum = temp
24
25            min_num = min(min_num ,postNum)
26
27            if ((curX == x2 and curY == y1) or
28                (curX == x2 and curY == y2) or
29                (curX == x1 and curY == y2)):
30                direction += 1
31            
32            if curX == x1 and curY == y1:
33                break
34
35        answer.append(min_num)
36
37    return answer