给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。
示例 1:
输入: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] 输出: [1,2,3,6,9,8,7,4,5]示例 2:
输入: [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] 输出: [1,2,3,4,8,12,11,10,9,5,6,7] class Solution: def spiralOrder(self, matrix: 'List[List[int]]') -> 'List[int]': if len(matrix) == 0: return [] up = 0 down = len(matrix)-1 left = 0 right = len(matrix[0])-1 res = list() while up <= down and left <= right: # 上 res.extend(matrix[up][left:right+1]) up += 1 if up > down: break # 右 # i = up # while i <= down: # res.append(matrix[i][right]) # i += 1 res.extend([matrix[i][right] for i in range(up, down + 1)]) # 替代上面的循环 right -= 1 if left > right: break # 下 temp = matrix[down][left:right+1] res.extend(temp[::-1]) # 逆序 down -= 1 if up > down: break # 左 # i = down # while i >= up: # res.append(matrix[i][left]) # i -= 1 res.extend([matrix[i][left] for i in range(down, up-1, -1)]) # 替代上面的循环,但执行用时似乎比上面的长 left += 1 if left > right: break return res if __name__=='__main__': s = Solution() matrix = \ [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12] ] print(s.spiralOrder(matrix))
