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.

  

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Knuth–Morris–Pratt string search algorithm

Start at LHS of string, string[0], trying to match pattern, working right. 
Trying to match string[i] == pattern[j].

 

Given a search pattern, pre-build a table, next[j], showing, when there is a mismatch at pattern position j, where to reset j to. 

If match fails, keep i same, reset j to position next[j].

 

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How to build the table

Everything else below is just how to build the table.

Construct a table showing where to reset j to

1. If mismatch string[i] != pattern[0], just move string to i+1, j = 0
2. If mismatch string[i] != pattern[1], we leave i the same, j = 0

   pattern = 10 
   string = ... 1100000

3. If mismatch string[i] != pattern[2], we leave i the same, and change j, but we need to consider repeats in pattern[0] .. pattern[1]

   pattern = 110 
   string = ... 11100000 
   i stays same, j goes from 2 back to 1

   pattern = 100 
   string = ... 10100000 
   i stays same, j goes from 2 back to 0

4. If mismatch string[i] != pattern[j], we leave i the same, and change j, but we need to consider repeats in pattern[0] .. pattern[j-1]

Given a certain pattern, construct a table showing where to reset j to.

 

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Construct a table of next[j]

For each j, figure out: 
next[j] = length of longest prefix in "pattern[0] .. pattern[j-1]" that matches the suffix of "pattern[1] .. pattern[j]” 

next[j] = “最大匹配的子串的长度"  
That is:

1. prefix must include pattern[0]
2. suffix must include pattern[j]
3. prefix and suffix are different

key

                                                        

Example for pattern  “ABABAC":

 

next[j] = length of longest prefix in "pattern[0] .. pattern[j-1]" that matches the suffix of "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
substring 0 to j	A	AB	ABA	ABAB	ABABA	ABABAC
longest prefix-suffix match	none	none	A	AB	ABA	none
next[j]	0	0	1	2	3	0
notes	no prefix and suffix that are different  i.e. next[0]=0 for all patterns

Given j, let n = 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]"

If match fails at position j+1(compare with s[j+1]), keep i same, reset pattern to position n(next[j]). 
Have already matched pattern[0] .. pattern[n-1],    pattern[0] .. pattern[n-1]=pattern[1] .. pattern[n]

e.g. We have matched ABABA so far. 
If next one fails, say we have matched ABA so far and then see if next one matches. 
That is, keep i same, just reset j to 3 (= precisely length of longest prefix-suffix match) 
Then, if match after ABA fails too, by the same rule we say we have matched A so far, reset to j = 1, and try again from there. 
In other words, it starts by trying to match the longest prefix-suffix, but if that fails it works down to the shorter ones until exhausted (no prefix-suffix matches left).

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Algorithm to construct table of next[j]

Do this once, when the pattern comes in.
pattern[0] ... pattern[m-1] 
Here, i and j both index pattern.

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 )
{

  // on 1 step i=1,j=0 
  if ( pattern[j] == pattern[i] )
  {
    next[i] = j+1 // it’s i not j
    i++
    j++
  }
  else ( pattern[j] != pattern[i] )
  {
    if ( j > 0 ){

            // 比如[0],[1],[2]  === [4],[5][6]

            //  这时 [3] <> [7]

     //maybe there is another pattern we can shift right though,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, just to make it clear what we are looping with
     }
  }
}

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