在leetcode中,我有一个问题来检查一系列无序的字符串“U”,“D”,“L”,“R”是否会形成一个圆圈。
我的提交是这样的:
def judgeCircle(moves):
l=r=u=d=0
for i in moves:
if i == 'L':
l+=1
if i == 'D':
d+=1
if i == 'R':
r+=1
if i == 'U':
u+=1
return ((l-r)==0) and ((u-d)==0)
而评委认为它耗时 239 毫秒 而另一个单行解决方案:
def judgeCircle(moves):
return (moves.count('R')==moves.count('L')) and
(moves.count('U')==moves.count('D'))
仅需 39 毫秒?
虽然我明白代码越少越好,但是我以为2nd会循环4次,我是不是误会了?
谢谢
最佳答案
这里有一些 timeit
代码显示了各种方法的速度,使用了所有 4 个键的数量相等的完美数据,以及每个键的数量大致相等的随机数据。
#!/usr/bin/env python3
''' Test speeds of various algorithms that check
if a sequence of U, D, L, R moves make a closed circle.
See https://stackoverflow.com/q/46568696/4014959
Written by PM 2Ring 2017.10.05
'''
from timeit import Timer
from random import seed, choice, shuffle
from collections import Counter, defaultdict
def judge_JH0(moves):
l = r = u = d = 0
for i in moves:
if i == 'L':
l += 1
if i == 'D':
d += 1
if i == 'R':
r += 1
if i == 'U':
u += 1
return ((l-r) == 0) and ((u-d) == 0)
def judge_JH1(moves):
l = r = u = d = 0
for i in moves:
if i == 'L':
l += 1
elif i == 'D':
d += 1
elif i == 'R':
r += 1
elif i == 'U':
u += 1
return (l == r) and (u == d)
def judge_count(moves):
return ((moves.count('R') == moves.count('L')) and
(moves.count('U') == moves.count('D')))
def judge_counter(moves):
d = Counter(moves)
return (d['R'] == d['L']) and (d['U'] == d['D'])
def judge_dict(moves):
d = {}
for c in moves:
d[c] = d.get(c, 0) + 1
return ((d.get('R', 0) == d.get('L', 0)) and
(d.get('U', 0) == d.get('D', 0)))
def judge_defdict(moves):
d = defaultdict(int)
for c in moves:
d[c] += 1
return (d['R'] == d['L']) and (d['U'] == d['D'])
# All the functions
funcs = (
judge_JH0,
judge_JH1,
judge_count,
judge_counter,
judge_dict,
judge_defdict,
)
def verify(data):
print('Verifying...')
for func in funcs:
name = func.__name__
result = func(data)
print('{:20} : {}'.format(name, result))
print()
def time_test(data, loops=100):
timings = []
for func in funcs:
t = Timer(lambda: func(data))
result = sorted(t.repeat(3, loops))
timings.append((result, func.__name__))
timings.sort()
for result, name in timings:
print('{:20} : {}'.format(name, result))
print()
# Make some data
keys = 'DLRU'
seed(42)
size = 100
perfect_data = list(keys * size)
shuffle(perfect_data)
print('Perfect')
verify(perfect_data)
random_data = [choice(keys) for _ in range(4 * size)]
print('Random data stats:')
for k in keys:
print(k, random_data.count(k))
print()
verify(random_data)
loops = 1000
print('Testing perfect_data')
time_test(perfect_data, loops=loops)
print('Testing random_data')
time_test(random_data, loops=loops)
典型输出
Perfect
Verifying...
judge_JH0 : True
judge_JH1 : True
judge_count : True
judge_counter : True
judge_dict : True
judge_defdict : True
Random data stats:
D 89
L 100
R 101
U 110
Verifying...
judge_JH0 : False
judge_JH1 : False
judge_count : False
judge_counter : False
judge_dict : False
judge_defdict : False
Testing perfect_data
judge_counter : [0.11746118000155548, 0.11771785900054965, 0.12218693499744404]
judge_count : [0.12314812499971595, 0.12353860199800692, 0.12495016200409736]
judge_defdict : [0.20643479600403225, 0.2069275510002626, 0.20834802299941657]
judge_JH1 : [0.25801684000180103, 0.2689959089984768, 0.27642749399819877]
judge_JH0 : [0.36819701099739177, 0.37400564400013536, 0.40291943999909563]
judge_dict : [0.3991459790049703, 0.4004156189985224, 0.4040740730051766]
Testing random_data
judge_count : [0.061543637995782774, 0.06157537500257604, 0.06704995800100733]
judge_counter : [0.11995147699781228, 0.12068584300141083, 0.1207217440023669]
judge_defdict : [0.2096717179956613, 0.21544414199888706, 0.220649760995002]
judge_JH1 : [0.261116588000732, 0.26281095200101845, 0.2706491360004293]
judge_JH0 : [0.38465088899829425, 0.38476935599464923, 0.3921787180006504]
judge_dict : [0.40892754300148226, 0.4094729179996648, 0.4135226650032564]
这些时间是在我的 Linux 上运行 Python 3.6.0 的旧 2GHz 32 位机器上获得的。
这里还有几个函数。
def judge_defdictlist(moves):
d = defaultdict(list)
for c in moves:
d[c].append(c)
return (len(d['R']) == len(d['L'])) and (len(d['U']) == len(d['D']))
# Sort to groups in alphabetical order: DLRU
def judge_sort(moves):
counts = [sum(1 for _ in g) for k, g in groupby(sorted(moves))]
return (counts[0] == counts[3]) and (counts[1] == counts[2])
judge_defdictlist
比 judge_defdict
慢但比 judge_JH1
快,当然它比 judge_defdict
使用更多的 RAM .
judge_sort
比 judge_JH0
慢,但比 judge_dict
快。
关于python - 在 python 中,为什么 string.count() 比循环更快?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46568696/