leetcode 54. 螺旋矩阵

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解题思路

模拟

时间复杂度

O(mn)

空间复杂度

O(mn)

代码

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class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        ret = []
        col = len(matrix[0])
        row = len(matrix)
        visited = [[False]*col for i in range(row)]
        x = 0
        y = 0
        direct = 'right'

        while True:
            if x >= row or y >= col or visited[x][y]:
                break
            
            ret.append(matrix[x][y])
            visited[x][y] = True

            if direct == 'right':
                if y == col - 1 or visited[x][y+1]:
                    direct = 'down'
                    x += 1
                else:
                    y += 1
                continue

            if direct == 'down':
                if x == row - 1 or visited[x+1][y]:
                    direct = 'left'
                    y -= 1
                else:
                    x += 1
                continue
            
            if direct == 'left':
                if y == 0 or visited[x][y-1]:
                    direct = 'up'
                    x -= 1
                else:
                    y -= 1
                continue

            if direct == 'up':
                if x == 0 or visited[x-1][y]:
                    direct = 'right'
                    y += 1
                else:
                    x -= 1
                continue

        return ret