Here's my problem:
Given a list of numbers and the number 'k', return whether any two numbers from the list add up to 'k'
For example, given [1,2,5,6]
where k
is 7, return True
since 2+5 is 7.
Here's my code; I'd like some help on how to vectorize it.
L = [3,4,1]
k = 5
for i in L:
for j in L:
if i+j == k:
print("True")
break
if i+j == k:
break
Vectorization doesn't seem like it would help here.
However, I propose a new algorithm that reduces complexity to O(N) from O(N^2):
L = [3,4,1]
Lset = set(L)
k = 5
for x in L:
if k - x in Lset:
print("True")
break
I think the algorithm is fairly self-explanatory! :) It basically iterates through L
and, for each number, checks if its "complimentary" (the number required to satisfy x + complimentary == k
) is present in the list (converted to a set for lower membership check cost).
If you don't want to count cases like L = [1]
, k = 2
, you can do this:
for x in L:
if k - x in Lset:
if k - x == x and L.count(x) < 2:
continue
print("True")
break
使用 itertools.combinations(L, 2) 为您提供所有 2 个长度的序列
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.