[Leetcode] 54. Spiral Matrix

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

 

 

Solution:

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        result = []
        rows, columns = len(matrix), len(matrix[0])
        left = up = 0
        right, down = columns-1, rows-1
        
        
        while len(result) < rows * columns:
            # Traverse from left to right.
            for col in range(left, right + 1):
                result.append(matrix[up][col])
            
            # up to down
            for row in range(up+1,down+1):
                result.append(matrix[row][right])
            
            # right to left
            if up != down:
                for col in range(right-1,left-1,-1):
                    result.append(matrix[down][col])

            if left != right:
                # Traverse upwards.
                for row in range(down - 1, up, -1):
                    result.append(matrix[row][left])
                    
            left += 1
            right -= 1
            up += 1
            down -= 1
            
        return result