Question

我正在https://www.hackerrank.com/challenges/angry-children处理一个问题。我写了一个正确解决一些测试用例的解决方案。其他测试用例，提供更多输入，超时的情况。如何更改此代码以更快地处理？

N = int(input())
K = int(input())
D = K - 1
N_set = []
for n in range(N):
    N_set.append(int(input()))
N_set.sort(reverse=True)

#Find differences between each integer in the list
D_set = []
for d in range(0,N-1):
    D_set.append(N_set[d]-N_set[d+1])

D_Start = 0
D_End = D
min_summed_diff = 99999999999999
D_Start_Hold = None
D_End_Hold = None

count_down = len(D_set) - D + 1
while count_down:
    #print(count_down)
    temp_summed_diff = 0
    for i in range(D_Start, D_End):
        temp_summed_diff += D_set[i]

    if temp_summed_diff < min_summed_diff:
        min_summed_diff = temp_summed_diff
        D_Start_Hold = D_Start
        D_End_Hold = D_End

    D_Start += 1
    D_End += 1
    count_down -= 1

K_set = N_set[D_Start_Hold:D_End_Hold+1]
unfairness = max(K_set) - min(K_set)
print(unfairness)

Answer 1

两个初始循环将更快，方法循环被拉出循环。列表理解也更快。我跑了这个实验

from timeit import timeit

a = '''\
tem = []
for i in range(%d):
  tem.append(int('1000'))
'''

b = '''\
tem = []
temapp = tem.append
for i in range(%d):
  temapp(int('1000'))
'''

c = "tem = [int('1000') for i in range(%d)]"

for n in (10,1000, 1000000):
    number = 1000000 // n
    print(timeit(a%n, number = number))
    print(timeit(b%n, number = number))
    print(timeit(c%n, number = number))
    print()

有了这些结果。

0.48560521545219426
0.3943799918166562
0.4146867965037263

0.372683063890642
0.3126639858171769
0.2900946187597775

0.36497313668305287
0.33567233916055894
0.3071572902003692

我相信这笔钱

    temp_summed_diff = 0
    for i in range(D_Start, D_End):
        temp_summed_diff += D_set[i]

就是这个，会更快

    temp_summed_diff = sum(D_set[D_start:D_end]

我没有看到任何其他加速。

Answer 2

您可以大大提高计算最小不公平性的算法。

如果你的列表被排序，那么每个子列表的不公平性只是它的结束之间的差异，所以这一切都归结为找到所有长度为k的子列表的最小差异。

此外，在这种情况下，通常可以更快地避免input()并直接从sys.stdin阅读。

这整个问题只需6行python即可解决：

import sys
nrs = map(int, sys.stdin)
n, k = next(nrs), next(nrs)
nrs = sorted(nrs)
min_unfair = min(nrs[k+i-1] - nrs[i] for i in range(0, len(nrs)-k+1))
print(min_unfair)

更高效的Python 3

2 个答案: