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[英]What's the time complexity of str.count and str.index in a for loop
[英]python str.index time complexity
为了找到字符串内子字符串的位置,幼稚算法将花费O(n^2)
时间。 但是,使用一些有效的算法(例如KMP算法 ),可以在O(n)时间内实现:
s = 'saurabh'
w = 'au'
def get_table():
i = 0; j = 2
t = []
t.append(-1); t.append(0)
while i < len(w):
if w[i] == w[j-1]:
t.append(j+1)
j += 1
else:
t.append(0)
j = 0
i += 1
return t
def get_word():
t = get_table()
i = j = 0
while i+j < len(s):
if w[j] == s[i+j]:
if j == len(w) - 1:
return i
j += 1
else:
if t[j] > -1:
i = i + j - t[j]
j = t[j]
else:
i += 1
return -1
if __name__ == '__main__':
print get_word()
但是,如果我们这样做: 'saurabh'.index('ra')
,它是在内部使用某种高效算法在O(n)
进行计算还是使用复杂度为O(n^2)
朴素算法?
在那篇文章[1]中,作者详细介绍了算法。 来自文章:
The function “fastsearch” is called. It is a mix between
Boyer-Moore and Horspool algorithms plus couple of neat tricks.
并从Boyer–Moore–Horspool算法的Wiki页面[2]:
The algorithm trades space for time in order to obtain an
average-case complexity of O(N) on random text, although
it has O(MN) in the worst case, where the length of the
pattern is M and the length of the search string is N.
希望有帮助!
[1] http://www.laurentluce.com/posts/python-string-objects-implementation
[2] https://en.wikipedia.org/wiki/Boyer%E2%80%93Moore%E2%80%93Horspool_algorithm
有时候,只要尝试一下就可以快速得到答案
>>> timeit.timeit('x.index("ra")', setup='x="a"*100+"ra"')
0.4658635418727499
>>> timeit.timeit('x.index("ra")', setup='x="a"*200+"ra"')
0.7199222409243475
>>> timeit.timeit('x.index("ra")', setup='x="a"*300+"ra"')
0.9555441829046458
>>> timeit.timeit('x.index("ra")', setup='x="a"*400+"ra"')
1.1994099491303132
>>> timeit.timeit('x.index("ra")', setup='x="a"*500+"ra"')
1.4850994427915793
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