Spiral Matrix
Given an m x n matrix, return all elements of the matrix in spiral order.
Constraints:
- m == matrix.length
- n == matrix[i].length
- 1 <= m, n <= 10
- -100 <= matrix[i][j] <= 100
Examples:
Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Explanation: The spiral order is: 1 -> 2 -> 3 -> 6 -> 9 -> 8 -> 7 -> 4 -> 5
Solutions
Directional Approach
Time: O(m*n)Space: O(1)
The solution uses four pointers (top, bottom, left, right) to represent the current boundaries of the matrix. It iterates through the matrix in a spiral order by moving the pointers accordingly.
def spiralOrder(matrix):
result = []
top, bottom, left, right = 0, len(matrix) - 1, 0, len(matrix[0]) - 1
while top <= bottom and left <= right:
for i in range(left, right + 1):
result.append(matrix[top][i])
top += 1
for i in range(top, bottom + 1):
result.append(matrix[i][right])
right -= 1
if top <= bottom:
for i in range(right, left - 1, -1):
result.append(matrix[bottom][i])
bottom -= 1
if left <= right:
for i in range(bottom, top - 1, -1):
result.append(matrix[i][left])
left += 1
return resultFollow-up:
What if the matrix is not a square matrix?
