LeetCode：蛇形矩阵II（spiral matrix II）

题目：

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

For example,
Given n = 3,

You should return the following matrix:

[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]

思路：

1. 判断特殊情况，n=0或1

2. 定义左右上下边界，依次填充数组；动态更新边界，继续填充，直到左右边界重合或上下边界重合。

3. 注意每次填充的前提是num小于n*n

代码（Python）：时间 O(n),  空间 O(n) 

class Solution(object):
def generateMatrix(self, n):
"""
:type n: int
:rtype: List[List[int]]
"""
if n == 0: return []
if n == 1: return [[1]]
arr = [[0]*n for i in range(n)]
row_begin = 0; row_end = n-1; col_begin = 0; col_end = n-1
num = 1; final = n*n
while num <= final:
if num <= final and col_begin < col_end:
for i in range(col_begin, col_end+1):
arr[row_begin][i] = num
num += 1
row_begin += 1
if num <= final and row_begin < row_end:
for i in range(row_begin, row_end+1):
arr[i][col_end] = num
num += 1
col_end -= 1
if num <= final and col_begin < col_end:
for i in range(col_end, col_begin-1,-1):
arr[row_end][i] = num
num += 1
row_end -= 1
if num <= final and row_begin < row_end:
for i in range(row_end, row_begin-1,-1):
arr[i][col_begin] = num
num += 1
col_begin += 1
return arr

简化后的代码：只用两个变量作为边界判断条件。

class Solution(object):
def generateMatrix(self, n):
"""
:type n: int
:rtype: List[List[int]]
"""
if n == 0: return []
arr = [[0]*n for i in range(n)]
start = 0; end = n-1; num = 1; final = n*n
while start < end:
for j in range(start, end+1):
arr[start][j] = num
num += 1
for i in range(start+1, end+1):
arr[i][end] = num
num += 1
for j in range(end-1, start-1,-1):
arr[end][j] = num
num += 1
for i in range(end-1, start,-1):
arr[i][start] = num
num += 1
start += 1
end -= 1
if start == end: arr[start][end]=final
return arr