彻底弄明白之数据结构中的KMP算法

如何加速朴素查找算法?KMP,当然还有其他算法,后续介绍.

Knuth–Morris–Prattstringsearchalgorithm

StartatLHSofstring,string[0],tryingtomatchpattern,workingright.
Tryingtomatchstring[i]==pattern[j].

Givenasearchpattern,pre-buildatable,next[j],showing,whenthereisamismatchatpatternpositionj,wheretoresetjto.

Ifmatchfails,keepisame,resetjtopositionnext[j].

Howtobuildthetable

Everythingelsebelowisjusthowtobuildthetable.

Constructatableshowingwheretoresetjto

Ifmismatchstring[i]!=pattern[0],justmovestringtoi+1,j=0

Ifmismatchstring[i]!=pattern[1],weleaveithesame,j=0
pattern=10
string=...1100000

Ifmismatchstring[i]!=pattern[2],weleaveithesame,andchangej,butweneedtoconsiderrepeatsinpattern[0]..pattern[1]
pattern=110
string=...11100000
istayssame,jgoesfrom2backto1

pattern=100
string=...10100000
istayssame,jgoesfrom2backto0

Ifmismatchstring[i]!=pattern[j],weleaveithesame,andchangej,butweneedtoconsiderrepeatsinpattern[0]..pattern[j-1]

Givenacertainpattern,constructatableshowingwheretoresetjto.

Constructatableofnext[j]

Foreachj,figureout:
next[j]=lengthoflongestprefixin"pattern[0]..pattern[j-1]"thatmatchesthesuffixof"pattern[1]..pattern[j]”
next[j]=“最大匹配的子串的长度"
Thatis:
prefixmustincludepattern[0]

suffixmustincludepattern[j]

prefixandsuffixaredifferent

key

Exampleforpattern“ABABAC":

next[j]=lengthoflongestprefixin"pattern[0]..pattern[j-1]"thatmatchesthesuffixof"pattern[1]..pattern[j]”

当j+1位与s[k]位比较,不匹配时

j'=next[j],j’和s[k]比较了,j’移到了原j+1的位置

j	0	1	2	3	4	5
substring0toj	A	AB	ABA	ABAB	ABABA	ABABAC
longestprefix-suffixmatch	none	none	A	AB	ABA	none
next[j]	0	0	1	2	3	0
notes	noprefixandsuffixthataredifferent i.e.next[0]=0forallpatterns

Givenj,letn=next[j]
"pattern[0]..pattern[n-1]"="pattern[j-(n-1)]..pattern[j]"

"pattern[0]..pattern[next[j]-1]"="pattern[j-(next[j]-1)]..pattern[j]"

e.g.j=4,n=3,

"pattern[0]..pattern[2]"="pattern[2]..pattern[4]"

Ifmatchfailsatpositionj+1(comparewiths[j+1]),keepisame,resetpatterntopositionn(next[j]).
Havealreadymatchedpattern[0]..pattern[n-1],pattern[0]..pattern[n-1]=pattern[1]..pattern

e.g.WehavematchedABABAsofar.
Ifnextonefails,saywehavematchedABAsofarandthenseeifnextonematches.
Thatis,keepisame,justresetjto3(=preciselylengthoflongestprefix-suffixmatch)
Then,ifmatchafterABAfailstoo,bythesamerulewesaywehavematchedAsofar,resettoj=1,andtryagainfromthere.
Inotherwords,itstartsbytryingtomatchthelongestprefix-suffix,butifthatfailsitworksdowntotheshorteronesuntilexhausted(noprefix-suffixmatchesleft).

Algorithmtoconstructtableofnext[j]

Dothisonce,whenthepatterncomesin.
pattern[0]...pattern[m-1]
Here,iandjbothindexpattern.
就是说是两个模式串在比较

next[0]=0 i=1 j=0 m=pattern.length while(i<m) { //on1stepi=1,j=0 if(pattern[j]==pattern[i]) { next[i]=j+1//it’sinotj i++ j++ } else(pattern[j]!=pattern[i]) { if(j>0){ //比如[0],[1],[2]===[4],[5][6] //这时[3]<>[7] //maybethereisanotherpatternwecanshiftrightthough,就是前缀和后缀 j=next[j-1]//因为next[j]就是给j+1用的,这个可记为定律,并且用j-1的原因还有0到[j-1]才有前后缀匹配的概念, //j是没有和模式串中的前缀匹配的,画画图就知道了 } else(j==0) { //模式串的下标为0时,与文本串s的下标i的值不匹配,i右移一位,模式串右移一位,0右移还是0 next[i]=0 i++ j=0//redundant,justtomakeitclearwhatweareloopingwith } } }