Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

Example:

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

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之前做过一个旋转数组的题，当时是旋转打印矩阵，这次只是构造一个旋转数组。方案还是一样的，条件变量也一样。时间复杂度为O(n*n),  空间复杂度为O(n)。 解决代码

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1     def generateMatrix(self, n): 2         """ 3         :type n: int 4         :rtype: List[List[int]] 5         """ 6         if n == 0 or n == 1: 7             return 0 if n == 0 else [[1]] 10         nums = 1 11         up, down, left, right = 0, n-1, 0, n-1   # 上下左右变量12         res = []13         for i in range(n):       # 结果矩阵14             res.append([0]*n)15             16         while left < right:     # 循环条件，也可以是 up< down。 因为是n*n矩阵，所以可以这样。17             for i in range(left, right):  18                 res[up][i] = nums19                 nums+= 120             21             for i in range(up, down):22                 res[i][right] = nums23                 nums += 124             25             for i in range(right, left, -1):26                 res[down][i] = nums27                 nums += 128             29             for i in range(down, up, -1):30                 res[i][left] = nums31                 nums += 132             up, down, left, right = up+1, down-1, left+1, right-133         34         if n % 2:                 # 如果是奇数的话，中心还有一个元素。35             res[left][right] = nums36         return res