Leetcode # 54. Spiral Matrix

https://leetcode.com/problems/spiral-matrix

Solution

Time Complexity: O(len(matrix) * len(matrix[0]))
Space Complexity: O(len(matrix) * len(matrix[0]))
(The input and output generally do not count towards the space complexity.)

class Solution:
  def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
    def _spiralOrder(matrix, start, size):
      if size[0] < 1 or size[1] < 1:
        return []
      spiral_order = []
      for j in range(size[1]):
        x = start[0] + j
        y = start[1]
        spiral_order.append(matrix[y][x])
      for i in range(size[0] - 1):
        x = start[0] + size[1] - 1
        y = start[1] + 1 + i
        spiral_order.append(matrix[y][x])
      if size[0] > 1 and size[1] > 1:
        for j in range(size[1] - 1):
          x = start[0] + size[1] - 2 - j
          y = start[1] + size[0] - 1
          spiral_order.append(matrix[y][x])
        for i in range(size[0] - 2):
          x = start[0]
          y = start[1] + size[0] - 2 - i
          spiral_order.append(matrix[y][x])
      return spiral_order + \
             _spiralOrder(matrix, 
                          (start[0] + 1, start[1] + 1), 
                          ( size[0] - 2,  size[1] - 2))

    return _spiralOrder(matrix, (0, 0), (len(matrix), len(matrix[0])))

Last Updated on 2023/08/16 by A1go