最大化列表中相邻元素之间差异的总和 [Python]

Question

问题陈述：一个列表中有N个元素，排列方式使相邻元素的绝对差值乘以索引的和最大，

索引从 1 开始。

(i=2,n=(elements in list))Σ i*abs(list[i-1]-list[i-2])

例子：

输入列表 :2 5 4

输出：13

解释：

[2,4,5] -> 2*(2) + 3*(1) = 7

[2,5,4] -> 2*(3) + 3*(1) = 9

[5,4,2] -> 2*(1) + 3*(2) = 8

[5,2,4] -> 2*(3) + 3*(2) = 12

[4,5,2] -> 2*(1) + 3*(3) = 11

[4,2,5] -> 2*(2) + 3*(3) = 13

Answer 1

您可能不会直接在这个论坛上写下问题并期望人们解决它们。本网站用于修复尝试，而不是进行尝试。此外，这个地方不是数学问题，而是编程问题，更具体的代码。我建议你去https://math.stackexchange.com/ 询问数学问题。

但是，同样的规则适用于那里。您必须表明您对如何解决问题有一个想法，并且您至少已经尝试过。他们会给你概念知识，而不是解决问题。就像在线老师一样。

还有一件事，你知道 Python 是什么吗？既然你放了标签。

不过，我确实有办法解决您的问题。您需要先对列表进行排序（我会给您 Python 代码，并让您了解它在整个过程中所做的事情）。

numbers = sorted(numbers)

现在，我们的数字从头到尾排序。现在，我们必须成对地取谁的差异最大和谁的差异最小的数字（顺序无关紧要，因为它是绝对值）。这很容易，因为我们已将数字从大到小排序。我们需要做的就是从末尾挑选一个数字，然后从开头挑选一个，然后将它们都从列表中删除，以便我们下次有一个新的开头和结尾。像这样。

numbers = [1, 5, 4] #random collection of numbers; you can make it anything really
numbers = sorted(numbers)
import math
pairs = [] #A new list, collection, array, or whatever you want to call it
while len(numbers) == 0 or len(numbers) == 1: #Until there is one element or no elements left:
    pairs.insert(-1, [numbers[0], numbers[-1]]) #-1 means the first element going backwards. 0 means the first element going forwards. Please see below code for better explanation
    del numbers[0] #The first number from our original list of numbers is gone
    del numbers[-1] #The last number from our original list is gone


if pairs: #If our original list of numbers has a number
    pairs.insert(0, [numbers[0]]) #See below code for info on insert()
    del numbers[0]


for pair in pairs:
    numbers.extend(pair) #Instead of having groups of numbers, we now are putting them all into one list. Meaning, instead of something like [[5, 4], [4]] where [5,4] and [4] are separate groups, we have [5, 4, 4] where it is all one group

sum = 0 #The total sum in the end

for x in range(len(numbers) - 1, 1, -2): #Indexes in list start from 0. So to talk about the first number and the last number in a list with 5 numbers you would talk about the 0th element and 4th element (0, 1, 2, 3, 4). We are going from the last to the second element by twos (-2 signifies going backwards). 
    sum += (math.fabs(numbers[x] - numbers[x - 1])) * (x+1) #The formula you specified, keep in mind that our x is '2' in case of the '3rd' element, '0' in case of the first, and so on. math.fabs() is a function that does absolute value

print(numbers,sum, sep = "|||") #Numbers, remember was changed to be the best order it could be. I printed out the numbers, then a |||, then the sum

请安装 Python 3.4 运行此程序

---

我在程序中使用了几次“插入”命令。在insert(-1, [numbers[0], numbers[-1]] 中，我将[first_number, last_number] 插入到列表的最后一部分。这可能有点过头了，谷歌一下，你会明白的。反正我就是这样做的（1年前）：P。记住：数字从小到大排序。

Answer 2

解决它的一种方法是遍历所有排列并计算每个排列的总数，然后按如下方式对结果列表进行排序：

import itertools

input_list = [2, 5, 4]
result = []

for perm in itertools.permutations(input_list, len(input_list)):
    sum_diff = 0
    for i,pair in enumerate(itertools.izip(perm[:-1], perm[1:])):
        sum_diff += abs(pair[0]-pair[1]) * (i+2)
    result.append((sum_diff, perm))

print sorted(result)[-1]

给出结果：

(13, (4, 2, 5))

完整的result 列表是：

[(7, (2, 4, 5)), (8, (5, 4, 2)), (9, (2, 5, 4)), (11, (4, 5, 2)), (12, (5, 2, 4)), (13, (4, 2, 5))]

使用 Python 2.7 测试