这是一个使用itertools 的解决方案，几乎不需要数学，只是观察螺旋的样子。我认为它很优雅而且很容易理解。
from math import ceil, sqrt
from itertools import cycle, count, izip
def spiral_distances():
"""
Yields 1, 1, 2, 2, 3, 3, ...
"""
for distance in count(1):
for _ in (0, 1):
yield distance
def clockwise_directions():
"""
Yields right, down, left, up, right, down, left, up, right, ...
"""
left = (-1, 0)
right = (1, 0)
up = (0, -1)
down = (0, 1)
return cycle((right, down, left, up))
def spiral_movements():
"""
Yields each individual movement to make a spiral:
right, down, left, left, up, up, right, right, right, down, down, down, ...
"""
for distance, direction in izip(spiral_distances(), clockwise_directions()):
for _ in range(distance):
yield direction
def square(width):
"""
Returns a width x width 2D list filled with Nones
"""
return [[None] * width for _ in range(width)]
def spiral(inp):
width = int(ceil(sqrt(len(inp))))
result = square(width)
x = width // 2
y = width // 2
for value, movement in izip(inp, spiral_movements()):
result[y][x] = value
dx, dy = movement
x += dx
y += dy
return result
用法：
from pprint import pprint
pprint(spiral(range(1, 26)))
输出：
[[21, 22, 23, 24, 25],
[20, 7, 8, 9, 10],
[19, 6, 1, 2, 11],
[18, 5, 4, 3, 12],
[17, 16, 15, 14, 13]]
以下是缩短的相同解决方案：
def stretch(items, counts):
for item, count in izip(items, counts):
for _ in range(count):
yield item
def spiral(inp):
width = int(ceil(sqrt(len(inp))))
result = [[None] * width for _ in range(width)]
x = width // 2
y = width // 2
for value, (dx, dy) in izip(inp,
stretch(cycle([(1, 0), (0, 1), (-1, 0), (0, -1)]),
stretch(count(1),
repeat(2)))):
result[y][x] = value
x += dx
y += dy
return result
我忽略了您希望输入为 2D 数组这一事实，因为任何 1D 可迭代更有意义。如果需要，您可以轻松地展平输入二维数组。我还假设输出应该是一个正方形，因为我想不出你想要什么。如果正方形的长度是偶数并且输入太长，它可能会越过边缘并引发错误：同样，我不知道替代方案是什么。