[Leetcode] 0054. Spiral Matrix

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Given an m x n 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

內容目錄

Python

				
					# 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))

Python

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