📅  最后修改于: 2023-12-03 15:26:26.157000             🧑  作者: Mango
最大重叠字符串指的是在两个字符串中找到最长的相同部分并重叠在一起的字符串。这是一个常见的问题,在字符串匹配算法中有很多应用。
最简单的方法是使用两个嵌套的循环来比较两个字符串的所有可能的子串,然后找到最长的重叠子串。时间复杂度为$O(n^3)$,其中$n$是字符串的长度。
def max_overlap(s1, s2):
n1, n2 = len(s1), len(s2)
max_overlap = ''
for i in range(n1):
for j in range(n2):
k = 0
while i + k < n1 and j + k < n2 and s1[i+k] == s2[j+k]:
k += 1
if k > len(max_overlap):
max_overlap = s1[i:i+k]
return max_overlap
上述暴力算法的时间复杂度是比较高的,为了优化,我们可以使用KMP算法。
KMP算法是一种字符串匹配算法,它的核心思想是利用已知信息来避免无效的比较。具体的算法实现可以参考这里,这里只给出求最大重叠字符串的代码实现。
def kmp(s):
n = len(s)
pi = [0] * n
for i in range(1, n):
j = pi[i-1]
while j > 0 and s[i] != s[j]:
j = pi[j-1]
if s[i] == s[j]:
j += 1
pi[i] = j
return pi
def max_overlap(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
n1, n2 = len(s1), len(s2)
pi = kmp(s1)
j, max_overlap = 0, ''
for i in range(n2):
while j > 0 and s2[i] != s1[j]:
j = pi[j-1]
if s2[i] == s1[j]:
j += 1
if j > len(max_overlap):
max_overlap = s1[:j]
if j == n1:
break
return max_overlap
使用KMP算法求最大重叠字符串的时间复杂度为$O(n)$。
最大重叠字符串也可以使用后缀数组来求解。后缀数组是指将一个字符串的所有后缀按照字典序排序并存储在数组中。具体的算法实现可以参考这里,这里只给出求最大重叠字符串的代码实现。
def build_suffix_array(s):
n = len(s)
sa = list(range(n))
sa.sort(key=lambda i: s[i:])
return sa
def lcp(s, sa):
n = len(s)
k = 0
h = [0] * n
for i in range(n):
if sa[i] == 0:
h[i] = 0
continue
j = sa.index(sa[i]-1)
while i+k < n and j+k < n and s[i+k] == s[j+k]:
k += 1
h[i] = k
if k > 0:
k -= 1
return h
def max_overlap(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
n1, n2 = len(s1), len(s2)
s = s1 + '#' + s2
sa = build_suffix_array(s)
h = lcp(s, sa)
j, max_overlap = 0, ''
for i in range(1, n1+n2+1):
if (sa[i-1] < n1 and sa[i] > n1) or (sa[i-1] > n1 and sa[i] < n1):
if h[i] > len(max_overlap):
max_overlap = s[sa[i]:sa[i]+h[i]]
return max_overlap
使用后缀数组求最大重叠字符串的时间复杂度为$O(n\log n)$。
本文从暴力算法、KMP算法、后缀数组三个方面介绍了如何求解最大重叠字符串。这三种算法的时间复杂度分别为$O(n^3)$、$O(n)$和$O(n\log n)$,根据实际的应用场景选择相应的算法。