LeetCode Spiral Matrix

LeetCode Spiral Matrix
Spiral Matrix for solving LeetCode Problems

Original question

Returns the content of a matrix in a spiral output.

Note:

The matrix is not necessarily a square.

Example:

Input: matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]

Output: [1, 2, 3, 6, 9, 8, 7, 4, 5]

Solutions

Controls the border of the current traversal and inner indentation. Note that the following situations may occur when you scale down to the innermost part:

There is a value left in the center --- | 3 | --- multiple values in the center horizontal --------- | 3 4 5 6 | --------- multiple values in the center vertical --- | 2 | 3 | 4 | ---

Just process them separately.

AC Source Code

class Solution(object):    def spiralOrder(self, matrix):        """        :type matrix: List[List[int]]        :rtype: List[int]        """        if not matrix:            return []        left = top = 0        right = len(matrix[0]) - 1        bottom = len(matrix) - 1        result = []        while left < right and top < bottom:            for i in range(left, right):                result.append(matrix[top][i])            for i in range(top, bottom):                result.append(matrix[i][right])            for i in range(right, left, -1):                result.append(matrix[bottom][i])            for i in range(bottom, top, -1):                result.append(matrix[i][left])            left += 1            right -= 1            top += 1            bottom -= 1        if left == right and top == bottom:            result.append(matrix[top][left])        elif left == right:            for i in range(top, bottom + 1):                result.append(matrix[i][left])        elif top == bottom:            for i in range(left, right + 1):                result.append(matrix[top][i])        return resultif __name__ == "__main__":    assert Solution().spiralOrder([        [1, 2, 3],        [4, 5, 6],        [7, 8, 9]    ]) == [1, 2, 3, 6, 9, 8, 7, 4, 5]    assert Solution().spiralOrder([[2], [3]]) == [2, 3]    assert Solution().spiralOrder([[2, 3]]) == [2, 3]