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

Example:

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

   之前做過一個旋轉數組的題，當時是旋轉打印矩陣，這次只是構造一個旋轉數組。方案還是一樣的，條件變量也一樣。時間復雜度為O(n*n),  空間復雜度為O(n)。
解決代碼

 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   # 上下左右變量
12         res = []
13         for i in range(n):       # 結果矩陣
 
14             res.append([0]*n)
15             
16         while left < right:     # 循環條件，也可以是 up< down。 因為是n*n矩陣，所以可以這樣。
17             for i in range(left, right):  
18                 res[up][i] = nums
19                 nums+= 1
20             
21             for i in range(up, down):
 
22                 res[i][right] = nums
23                 nums += 1
24             
25             for i in range(right, left, -1):
26                 res[down][i] = nums
27                 nums += 1
28             
29             for i in range(down, up, -1):
30                 res[i][left] = nums
31                 nums += 1
32             up, down, left, right = up+1, down-1, left+1, right-1
33         
34         if n % 2:                 # 如果是奇數的話，中心還有一個元素。
35             res[left][right] = nums
36         return res

【LeetCode每天一題】Spiral Matrix II(螺旋數組II)