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54. Spiral Matrix

Difficulty Topics

Description

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

spiral-matrix.py
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        res = []
        rows, cols = len(matrix), len(matrix[0])
        rowStart, rowEnd = 0, rows - 1
        colStart, colEnd = 0, cols - 1

        while len(res) < rows * cols:
            # traverse right
            i = colStart
            while len(res) < rows * cols and i <= colEnd:
                res.append(matrix[rowStart][i])
                i += 1
            rowStart += 1

            # traverse down
            i = rowStart
            while len(res) < rows * cols and i <= rowEnd:
                res.append(matrix[i][colEnd])
                i += 1
            colEnd -= 1

            # traverse left
            i = colEnd
            while len(res) < rows * cols and i >= colStart:
                res.append(matrix[rowEnd][i])
                i -= 1
            rowEnd -= 1

            # traverse up
            i = rowEnd
            while len(res) < rows * cols and i >= rowStart:
                res.append(matrix[i][colStart])
                i -= 1
            colStart += 1


        return res