• 需求背景

需求背景

给你一个数组，把这个数组里面每个数字都求差，然后把这些差都求和。

正常演示步骤：

[1,3,5]
1-3 = 2
1-5 = 4
-
3-5 = 2

sum = 8

反过来思考子问题是什么：

[1,3,5]
reverse [5,3,1]

5-3 = 2
5-1 = 4
-
3-1 = 2

sum = 8

[1,3,5,7]
reverse [7,5,3,1]

7-5 = 2
7-3 = 4
7-1 = 6
-
5-3 = 2
5-1 = 4
-
3-1 = 2

sum = 20

写出代码：
code:

def f(n) -> int:
    temp = []

    if len(n) == 1:
        return sum(temp)

    reN = list(reversed(n))
    for i, x in enumerate(reN):  # i[5,3,1]
        if i != 0:
            print("reN ", reN)
            print(reN[0], reN[i])
            temp.append(abs(reN[0] - reN[i]))
            print(temp)
    return sum(temp) + f(n[:-1])


n = [1, 3, 5, 7]
print(f(n))