KMP algorithm for implementing string in Python

In this paper, we describe the KMP algorithm of Python implementation string. Share to everyone for your reference, as follows:

KMP Algorithm Python implementation

Today research KMP algorithm, it seems many versions, there are different languages written, but feel more chaotic, finally try to write a copy of the summary

First, the KMP algorithm makes the optimization algorithm in string matching, so that the original O (m*n) drops to O (m+n)

As for his understanding, I recommend watching the video first, he speaks it very clearly. Wang can understand. KMP String matching algorithm
Then from the visual aspects of understanding, recommended to see how to better understand and master the KMP algorithm? -The answer of the Blessed son
Finally understand searching for Patterns from the code layer | Set 2 (KMP algorithm)
Code do not see too many others, I feel a lot of people write also have problems, I run the problem, I have to see some students write, was taken to other pits ...
Last recorded code

"' first seek Next array ' Def next_list (pattern): next=[] pattern_list=list (pattern) j=0 I=1 for S in range (len (pattern ): #第一位一直是0 if S==0:next.append (0) #看第i个和第j个字母是否相同, if same, accumulate #同时i, J simultaneously right-move Eli F (Pattern_list[i]==pattern_list[j]): Next.append (j+1) j+=1 i+=1 #如果不相同, Then next is 0, and J also returns to the first position, I move forward a position else:next.append (0) j=0 i=s+1 print (next) RET Urn nextnext_list (' ababcabab ') def KMP (text,pattern): #text的位置 i=0 #pattern的位置 j=0 next=next_list (Patt     Ern) if (not (text and pattern)): print (' String null, enter String ') elif (len (text) <len (pattern)): Print (' Original string too small ') Elif (Text==pattern): print (' string consistent ') Else:while ((I<len (text))): Print ((text[i],patter                N[J]) print (i,j) #如果相同, then the next contrast if (Text[i]==pattern[j]): i+=1 J+=1 #判断Is it a match? If J==len (pattern): print (' start matching from {0} '. Format (I-J)) j=next[j-1] #如果不匹配, J back to the previous letter corresponding to the next elif I<len (text) and Text[i]!=pattern[j]: #我就是少了这个判断, there has been a problem, is not the horse After the second step, after this very critical if j!=0: #这个就是精髓了, if not match, go to find the first and this match the string, and then in this front of the matching string #的后一个字母拿                    Out, and then compared with the long text of the first letter of comparison j=next[j-1] #如果j已经回到了0, then by increasing the i,text to move to the next letter else: i+=1# Text = "Ababdabacdababcabab" # pattern = "Ababcabab" text= ' Abxabcabcaby ' pattern= ' abcaby ' km  P (text,pattern) #output: next=[0, 0, 0, 1, 2, 0] (' A ', ' a ') 0 0 (' b ', ' B ') 1 1 (' X ', ' a ') 2 0 (' A ', ' a ') 3 0 (' B ', ' B ') 4 1 (' C ', ' C ') 5 2 (' A ', ' a ') 6 3 (' B ', ' B ') 7 4 (' C ', ' C ') 8 2 (' A ', ' a ') 9 3 (' B ', ' B ') 4 (' Y ', ' y '), and the 6th position starts matching

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