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

Approach

Solution

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        res = []
        top, bottom = 0, len(matrix) - 1
        left, right = 0, len(matrix[0]) - 1

        while top <= bottom and left <= right:
            # 1. Top Row
            for c in range(left, right + 1):
                res.append(matrix[top][c])
            top += 1

            # 2. Right Column
            for r in range(top, bottom + 1):
                res.append(matrix[r][right])
            right -= 1

            # 3. Bottom Row
            if top <= bottom:
                for c in range(right, left - 1, -1):
                    res.append(matrix[bottom][c])
                bottom -= 1

            # 4. Left Column
            if left <= right:
                for r in range(bottom, top - 1, -1):
                    res.append(matrix[r][left])
                left += 1

        return res

Explanation