生成列表的最简单方法是：
def comparisons(seq):
return [abs(a-b) for a, b in zip(seq, seq[1:])]
至于你的比较,最高值总是最大值,然后是最小值,重复.例如：长度为4：
[3, 0, 3, 0]
因为这将每次产生最大差异.对于比较字符串(长度为1),每个项目将存在这些最大差异之一(长度为1).因此最大值将是(长度-1)** 2.
但是,你似乎暗示长度3的最大值是3,那么为什么[0,2,0]无效(产生[2,2],总和为4)？
您提到必须包含从0到长度为1的所有整数,但这会使您的一些示例(例如：[0,1,0])无效.这也与任何元素都可以重复的想法相冲突(如果长度为n的列表必须包含0到n-1,则它不能重复).
如果这种情况属实,那么您的问题就会变得与创建抖动矩阵的问题类似.
在排序范围从0到len-1的情况下,为了产生最大差异,最佳算法是从0开始,从len-1开始,将低值加到列表的最高“侧”. ,反之亦然：
from collections import deque
from itertools import permutations
from operator import itemgetter
def comparisons(seq):
return [abs(a-b) for a, b in zip(seq, seq[1:])]
def best_order(n):
temp = deque([0, n-1])
low = 1
high = n-2
while low < high:
left = temp[0]
right = temp[-1]
if left < right:
temp.append(low)
temp.appendleft(high)
else:
temp.append(high)
temp.appendleft(low)
low += 1
high -= 1
if len(temp) < n:
temp.append(low)
return list(temp)
def brute_force(n):
getcomp = itemgetter(2)
return max([(list(a), comparisons(a), sum(comparisons(a))) for a in permutations(range(n))], key=getcomp)
for i in range(2, 6):
print("Algorithmic:", best_order(i), comparisons(best_order(i)), sum(comparisons(best_order(i))))
print("Brute Force:", *brute_force(i))
这给了我们：
Algorithmic: [0, 1] [1] 1
Brute Force: [0, 1] [1] 1
Algorithmic: [0, 2, 1] [2, 1] 3
Brute Force: [0, 2, 1] [2, 1] 3
Algorithmic: [2, 0, 3, 1] [2, 3, 2] 7
Brute Force: [1, 3, 0, 2] [2, 3, 2] 7
Algorithmic: [3, 0, 4, 1, 2] [3, 4, 3, 1] 11
Brute Force: [1, 3, 0, 4, 2] [2, 3, 4, 2] 11
表明该算法与蛮力方法相匹配,可以产生最佳结果.
以下是一个更通用的解决方案：
from collections import deque
def comparisons(seq):
return [abs(a-b) for a, b in zip(seq, seq[1:])]
def best_order(seq):
pool = deque(sorted(seq))
temp = deque([pool.popleft(), pool.pop()])
try:
while pool:
if temp[0] < temp[-1]:
temp.append(pool.popleft())
temp.appendleft(pool.pop())
else:
temp.append(pool.pop())
temp.appendleft(pool.popleft())
except IndexError:
pass
return list(temp)
i = [0, 1, 2, 3, 4, 5, 6, 0, 0, 1, 1, 2, 2]
print("Algorithmic:", best_order(i), comparisons(best_order(i)), sum(comparisons(best_order(i))))
for n in range(2, 6):
i = list(range(n))
print("Algorithmic:", best_order(i), comparisons(best_order(i)), sum(comparisons(best_order(i))))
这使：
Algorithmic: [2, 1, 3, 0, 5, 0, 6, 0, 4, 1, 2, 1, 2] [1, 2, 3, 5, 5, 6, 6, 4, 3, 1, 1, 1] 38
Algorithmic: [0, 1] [1] 1
Algorithmic: [0, 2, 1] [2, 1] 3
Algorithmic: [2, 0, 3, 1] [2, 3, 2] 7
Algorithmic: [3, 0, 4, 1, 2] [3, 4, 3, 1] 11
这与之前的结果相符.
