递增数组的判断【python实现】

有时候需要对某一组数组的数据进行判断是否 递增 的场景，比如我在开发一些体育动作场景下，某些肢体动作是需要持续朝着垂直方向向上变化，那么z轴的值是会累增的。同理，逆向考虑，递减就是它的对立面。

下面是查找总结到的所有方式，如有补充可以评论区提出。

资料参考来源: Check if list is strictly increasing

1. zip() and all()

• Code:

test_list = [1, 4, 5, 7, 8, 10]
# Using zip() and all() to
# Check for strictly increasing list
res = all(i < j for i, j in zip(test_list, test_list[1:]))
print(f"Is list strictly increasing ? : {res}")

• Output:

Is list strictly increasing ? : True

时间复杂度: O(n), n是数组的长度。

2. reduce and lambda

• Code:

import functoolstest_list = [1, 4, 5, 7, 8, 10]
res = bool((lambda list_demo: functools.reduce(lambda i, j: j ifi < j else 9999, list_demo) != 9999)(test_list))print(f"Is list strictly increasing ? : {res}")

• Output:

Is list strictly increasing ? : True

时间复杂度: O(n), n是数组的长度。

3. itertools.starmap() + zip() + all()

• Code:

import itertoolstest_list = [1, 4, 5, 7, 8, 10]
res = all(itertools.starmap(operator.le, zip(test_list, test_list[1:])))print(f"Is list strictly increasing ? : {res}")

• Output:

Is list strictly increasing ? : True

时间复杂度: O(n), n是数组的长度。

4. sort() and extend()

• Code:

test_list = [1, 4, 5, 7, 8, 10]
res = False
new_list = []
new_list.extend(test_list)
test_list.sort()if new_list == test_list:res = Trueprint(f"Is list strictly increasing ? : {res}")

• Output:

Is list strictly increasing ? : True

时间复杂度: O(nlogn), 这里是sort()的时间复杂度

5. Use stacks

栈是一种后进先出的数据结构(Last in, first out)。

• Code:

def is_strictly_increasing(lst):stack = []for i in lst:if stack and i <= stack[-1]:return Falsestack.append(i)return Truetest_list = [1, 4, 5, 7, 8, 10]
print(is_strictly_increasing(test_list))  # Truetest_list = [1, 4, 5, 7, 7, 10]
print(is_strictly_increasing(test_list))  # False

时间复杂度: O(n)，原数组被遍历了一遍
空间复杂度: O(n)，栈可能要存储全部的n个原数组元素

6. numpy()

• Code:

import numpy as npdef is_increasing(lst):# Converting input list to a numpy arrayarr = np.array(lst)# calculate the difference between adjacent elements of the arraydiff = np.diff(arr)# check if all differences are positive# using the np.all() functionis_increasing = np.all(diff > 0)# return the resultreturn is_increasing# Input list
test_list = [1, 4, 5, 7, 8, 10]# Printing original lists
print("Original list : " + str(test_list))result = is_increasing(test_list)print(result)
# True

时间复杂度: O(n)

7. itertools.pairwise() and all()

这里面就等于使用 pairwise() 替代了之前的 zip(list, list[1:]) 。

• Code:

from itertools import pairwise# Function
def is_strictly_increasing(my_list):# using pairwise method to iterate through the list and# create pairs of adjacent elements.# all() method checks if all pairs of adjacent elements# satisfy the condition i < j, where i and j# are the two elements in the pair.if all(a < b for a, b in pairwise(my_list)):return Trueelse:return False# Initializing list
test_list = [1, 4, 5, 7, 8, 10]# Printing original lists
print("Original list : " + str(test_list))# Checking for strictly increasing list
# using itertools pairwise() and all() method
res = is_strictly_increasing(test_list)# Printing the result
print("Is list strictly increasing ? : " + str(res))

• Output:

Original list : [1, 4, 5, 7, 8, 10]
Is list strictly increasing ? : True

时间复杂度: O(n)