Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return [1,2,3,6,9,8,7,4,5]
.
class Solution: # @param matrix, a list of lists of integers # @return a list of integers def spiralOrder(self, matrix): m = len(matrix) if 0 == m: return [ ] n = len(matrix[0]) if 0 == n: return [ ] arr = [ ] round = (min(m, n) + 1) / 2 for x in range(0, round): for y in range(x, n - x): arr.append(matrix[x][y]) for y in range(x + 1, m - x - 1): arr.append(matrix[y][n - x - 1]) if m - 2 * x > 1: ### for y in range(n - x - 1, x - 1, -1): arr.append(matrix[m - x - 1][y]) if n - 2 * x > 1: ### for y in range(m - x - 2, x, -1): ### arr.append(matrix[y][x]) return arr