Question

在python中给我的问题是：接受一个具有n个术语的列表，并在不使用sort函数和def的情况下检查元素是否按升序，降序排列。

请帮助我执行此程序。

代码：

a=list(input("Enter elements:"))
for i in range(len(a)):
    for j in range(i):
        if a[j]<=a[j+1]:
            print("It is in ascending order")
        elif a[j]>=a[j+1]:
            print("It is in descending order")
        else:
            print("It is not in order")

Answer 1

def which_order(list1):
    isAscending = True
    isDescending = True
    for i in range(1,len(list1)):
        if(list1[i] >= list1[i-1]):
            isDescending = False
        elif(list1[i] <= list1[i-1]):
            isAscending = False
        if((not isAscending) and (not isDescending)):
            print("The list is not sorted in either order.")
            break
    if(isAscending):
        print("The list is sorted in ascending order.")
    if(isDescending):
        print("The list is sorted in descending order.")

示例输出：

list1 = [1,2,3,4]
list2 = [9,8,7,6]
list3 = [1,9,8,7,6]
which_order(list1)
which_order(list2)
which_order(list3)

The list is sorted in ascending order.
The list is sorted in descending order.
The list is not sorted in either order.

Answer 2

def func(l):
    a_cond = True # considering list is in ascending order
    d_cond = True # # considering list is in descending order

    # checking if for any number my ascending condition false

    for i in range(1,len(l)):
        if not l[i]>l[i-1]:
            a_cond = False
    # if condition not false then return list is ascending
    if a_cond:
        return 'Ascending Order'

    # if condition false then check for descending codition
    for i in range(1, len(l)):
        if not l[i]<l[i-1]:
            d_cond = False
    # check if all number in descending order
    if d_cond:
        return 'Descending order'

    # if above both condition fail, list is in mix order
    return 'Mix Order'



print(func([1,2,3,4,5]))
print(func([5,4,3,2,1]))
print(func([1,2,3,1,2]))

输出

Ascending Order
Descending order
Mix Order

如何在不使用排序功能的情况下查找列表中的元素是升序还是降序？

2 个答案: