需求背景

給你一個數組，把這個數組里面每個元素之間都求差（結果忽略負符號），然后把這些差求和。

給你數組 n = [1,3,5]
正常演示步驟：

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

sum = 8

反過來思考子問題是什么：

假設已經有一個x函數幫忙搞定了，自己之做一步：(3-1） + x()

[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

寫出代碼：
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("sum :",f(n))

out:

reN  [7, 5, 3, 1]
7  -  5
[2]
reN  [7, 5, 3, 1]
7  -  3
[2, 4]
reN  [7, 5, 3, 1]
7  -  1
[2, 4, 6]
reN  [5, 3, 1]
5  -  3
[2]
reN  [5, 3, 1]
5  -  1
[2, 4]
reN  [3, 1]
3  -  1
[2]
sum : 20