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Given an mxn
matrix
, return all elements of the matrix
in spiral order.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]] Output: [1,2,3,4,8,12,11,10,9,5,6,7]
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
内容目录
TogglePython
# time complexity: O(m*n) # space complexity: O(1) from typing import List class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: ROW = len (matrix) COL = len(matrix[0]) direction = 1 row = 0 col = -1 result = [] while ROW > 0 and COL > 0: for _ in range(COL): col += direction result.append (matrix[row][col]) ROW -= 1 for _ in range(ROW): row += direction result.append(matrix[row][col]) COL -= 1 direction *= -1 return result matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] print(Solution().spiralOrder(matrix)) matrix = [[1, 2, 3, 4], [ 5, 6, 7, 8], [9, 10, 11, 12]] print(Solution().spiralOrder(matrix))