84 lines
2.7 KiB
Python
84 lines
2.7 KiB
Python
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#!/usr/bin/python
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# -*- coding: UTF-8 -*-
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def kmp(main, pattern):
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"""
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kmp字符串匹配
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:param main:
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:param pattern:
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:return:
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"""
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assert type(main) is str and type(pattern) is str
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n, m = len(main), len(pattern)
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if m == 0:
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return 0
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if n <= m:
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return 0 if main == pattern else -1
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# 求解next数组
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next = get_next(pattern)
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j = 0
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for i in range(n):
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# 在pattern[:j]中，从长到短递归去找最长的和后缀子串匹配的前缀子串
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while j > 0 and main[i] != pattern[j]:
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j = next[j-1] + 1 # 如果next[j-1] = -1，则要从起始字符取匹配
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if main[i] == pattern[j]:
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if j == m-1:
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return i-m+1
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else:
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j += 1
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return -1
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def get_next(pattern):
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"""
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next数组生成
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注意：
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理解的难点在于next[i]根据next[0], next[1]…… next[i-1]的求解
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next[i]的值依赖于前面的next数组的值，求解思路：
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1. 首先取出前一个最长的匹配的前缀子串，其下标就是next[i-1]
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2. 对比下一个字符，如果匹配，直接赋值next[i]为next[i-1]+1，因为i-1的时候已经是最长
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*3. 如果不匹配，需要递归去找次长的匹配的前缀子串，这里难理解的就是递归地方式，next[i-1]
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是i-1的最长匹配前缀子串的下标结尾，则 *next[next[i-1]]* 是其次长匹配前缀子串的下标
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结尾
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*4. 递归的出口，就是在次长前缀子串的下一个字符和当前匹配 或 遇到-1，遇到-1则说明没找到任
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何匹配的前缀子串，这时需要找pattern的第一个字符对比
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ps: next[m-1]的数值其实没有任何意义，求解时可以不理。网上也有将next数组往右平移的做法。
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:param pattern:
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:return:
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"""
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m = len(pattern)
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next = [-1] * m
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next[0] = -1
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# for i in range(1, m):
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for i in range(1, m-1):
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j = next[i-1] # 取i-1时匹配到的最长前缀子串
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while j != -1 and pattern[j+1] != pattern[i]:
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j = next[j] # 次长的前缀子串的下标，即是next[next[i-1]]
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# 根据上面跳出while的条件，当j=-1时，需要比较pattern[0]和当前字符
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# 如果j!=-1，则pattern[j+1]和pattern[i]一定是相等的
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if pattern[j+1] == pattern[i]: # 如果接下来的字符也是匹配的，那i的最长前缀子串下标是next[i-1]+1
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j += 1
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next[i] = j
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return next
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if __name__ == '__main__':
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m_str = "aabbbbaaabbababbabbbabaaabb"
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p_str = "abbabbbabaa"
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print('--- search ---')
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print('[Built-in Functions] result:', m_str.find(p_str))
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print('[kmp] result:', kmp(m_str, p_str))
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