python实现leetcode之59. 螺旋矩阵 II

解题思路

在矩阵四周添加一圈障碍物
后续在碰到障碍物的时候调整方向
总共处理N * N个元素后停止

59. 螺旋矩阵 II

代码

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        matrix = [[None for _ in range(n)][:] for _ in range(n)]
        obstacle = set()
        for r in range(n):
            obstacle.add((r, -1))  # 每一行的开头加一个障碍物
            obstacle.add((r, n))   # 每一行的结尾加一个障碍物
            obstacle.add((-1, r))  # 每一列的开头加一个障碍物
            obstacle.add((n, r))   # 每一列的结尾加一个障碍物
        direction = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        count, total = 1, n * n
        di = 0
        x, y = 0, 0
        while count <= total:
            matrix[x][y] = count
            obstacle.add((x, y))
            next_x = x+direction[di][0]
            next_y = y+direction[di][1]
            if (next_x, next_y) in obstacle:
                di = (di+1) % 4
            x, y = x+direction[di][0], y+direction[di][1]
            count += 1
        return matrix