我在接受采访时被问到这个问题。鉴于以下列表:
[1,2,5,7,3,10,13]
找出最多相加的数字。
我的解决方案如下:
#sort the list:
l.sort()
result = ()
for i in range(len(l)):
for j in range(i, len(l)):
if l[i] + l[j] > 5:
break
elif l[i] + l[j] == 5:
result += (l[i], l[j])
我提出的想法是对列表进行排序,然后循环并查看总和是否大于5.如果是,那么我可以停止循环。我觉得面试官对这个答案不满意。有人可以为我将来的参考建议一个更好的吗?
答案 0 :(得分:3)
这将返回总和为5的输入的powerset的所有元素:
>>> input = [1,2,5,7,3,10,13]
>>> import itertools
>>> def powerset(l):
... return itertools.chain.from_iterable((itertools.combinations(l, i) for i in range(len(l)+1)))
...
>>> filter(lambda v: sum(v) == 5, powerset(input))
[(5,), (2, 3)]
答案 1 :(得分:2)
我就是这样做的:
from itertools import permutations
from random import randint
base_list = [randint(-10, 10) for x in range(20)]
def five_sum(base_list, combinations):
has_negatives = any([base for base in base_list if base < 0])
if not has_negatives:
filtered_list = [base for base in base_list if base <= 5]
else:
filtered_list = base_list
perms = list(permutations(filtered_list, combinations))
filtered_perms = [perm for perm in perms if sum(perm) == 5]
print(filtered_perms)
for perm in set(filtered_perms):
yield perm
print(base_list)
print(list(five_sum(base_list, 2)))
在我拥有无限内存的理想世界中,我会将combinations
参数替换为perms = [list(permutations(filtered_list, i)) for i in range(len(filtered_list) + 1)]
答案 2 :(得分:2)
使用词典的另一种解决方案
from collections import Counter
l = [1,2,2,5,7,3,10,0,-5]
counter = Counter(l)
keys = counter.keys()
result = []
key = keys.pop()
while True:
if 5-key in counter:
result.append((key, 5-key))
counter[key]-=1
if counter[key]<=0:
del counter[key]
if len(keys) == 0:
break
key = keys.pop()
counter[5-key]-=1
if counter[5-key]<=0:
del counter[5-key]
else:
del counter[key]
if len(keys) == 0:
break
key = keys.pop()
print(result)
你得到了
[(-5, 10), (5, 0), (3, 2)]
提案解决方案的
len(l)==1000
时间:
from timeit import Timer
t = Timer(jose)
print t.timeit(number=1)
#0.00108003616333
t = Timer(dan)
print t.timeit(number=1)
#hangout
t = Timer(morgan)
print t.timeit(number=1)
#0.000875949859619 <--- best time
t = Timer(steven)
print t.timeit(number=1)
#0.0118129253387
t = Timer(sam) #OP
print t.timeit(number=1)
#0.0160880088806