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

---

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

        while top <= bottom and left <= right:
            if direction == 0:
                for i in range(left, right+1):
                    result.append(matrix[top][i])
                top += 1
            elif direction == 1:
                for i in range(top, bottom+1):
                    result.append(matrix[i][right])
                right -= 1
            elif direction == 2:
                for i in range(right, left-1, -1):
                    result.append(matrix[bottom][i])
                bottom -= 1
            elif direction == 3:
                for i in range(bottom, top-1, -1):
                    result.append(matrix[i][left])
                left += 1
            direction = (direction+1) % 4

        return result