Python implements the KMP algorithm for strings

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Release： 2018-04-19 16:20:40

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The example in this article describes the KMP algorithm for string implementation in Python. Share it with everyone for your reference, the details are as follows:

KMP algorithm Python implementation

I studied the KMP algorithm today. It seems that there are many versions written in different languages. , but the more I looked at it, the more confusing it became. Finally, I tried to write a summary

First of all, the KMP algorithm is an optimization algorithm in string matching. The original O(m*n) has been reduced to O(m n)

Regarding his understanding, I recommend watching the video first, he explained it very clearly. The KMP string matching algorithm that Wang can understand
Then understand it from the visual aspect. It is recommended to see how to better understand and master the KMP algorithm? - Yuzi's answer-Zhihu Rong
Finally understand Searching from the code level for Patterns | Set 2 (KMP Algorithm)
Don’t read too much other people’s codes. I feel that many people have problems with their writing. When I ran into problems myself, I read some writing by some classmates and was taken to other pits. . . .
Last record code

&#39;&#39;&#39;
先求next数组
&#39;&#39;&#39;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，如果相同，则累加
        #同时i,j同时右移
        elif(pattern_list[i]==pattern_list[j]):           
            next.append(j+1)
            j+=1
            i+=1
        #如果不相同，则next为0，同时j也退回第一个位置，i继续前进一个位置
        else:
            next.append(0)
            j=0
            i=s+1
    print(next)    return next

next_list(&#39;ABABCABAB&#39;)      

def kmp(text,pattern):
    #text的位置
    i=0
    #pattern的位置
    j=0
    next=next_list(pattern)    if(not(text and pattern)):
        print(&#39;字符串为空，请输入字符串&#39;)    elif(len(text)<len(pattern)):
        print(&#39;原字符串过小&#39;)    elif(text==pattern):
        print(&#39;字符串一致&#39;)    else:        while( (i<len(text) )):
            print((text[i],pattern[j]))
            print(i,j)            #如果相同，则进行下一个对比
            if( text[i]==pattern[j]):
                i+=1
                j+=1
            #判断是不是匹配完了
            if j==len(pattern):
                print(&#39;从第{0}个位置开始匹配&#39;.format(i-j))
                j=next[j-1]            #如果不匹配，则j反回到前一个字母对应的next
            elif i<len(text) and text[i]!=pattern[j]:                #我就是少了这个判断，一直有问题，就是在不匹配后的第二步，后面这个很关键
                if j!=0:                #这个就是精髓了，如果不匹配，就去找第一个和这个匹配的字符串，然后在这个前面的匹配字符串
                #的后一个字母拿出来，再与长text比较的第i个字母比较
                    j=next[j-1]                #如果j已经回到了0，则通过增加i,text移动到下一个字母
                else:
                    i+=1# text = "ABABDABACDABABCABAB"# pattern = "ABABCABAB"            text=&#39;abxabcabcaby&#39;pattern=&#39;abcaby&#39;kmp(text,pattern)#output:next=[0, 0, 0, 1, 2, 0]


(&#39;a&#39;, &#39;a&#39;)
0 0
(&#39;b&#39;, &#39;b&#39;)
1 1
(&#39;x&#39;, &#39;a&#39;)
2 0
(&#39;a&#39;, &#39;a&#39;)
3 0
(&#39;b&#39;, &#39;b&#39;)
4 1
(&#39;c&#39;, &#39;c&#39;)
5 2
(&#39;a&#39;, &#39;a&#39;)
6 3
(&#39;b&#39;, &#39;b&#39;)
7 4
(&#39;c&#39;, &#39;c&#39;)
8 2
(&#39;a&#39;, &#39;a&#39;)
9 3
(&#39;b&#39;, &#39;b&#39;)
10 4
(&#39;y&#39;, &#39;y&#39;)
11 5
从第6个位置开始匹配

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