KMP模式匹配算法程序（Python，C++，C）

代码来自维基教科书：Knuth-Morris-Pratt pattern matcher。

Python程序如下：

# Knuth-Morris-Pratt string matching
# David Eppstein, UC Irvine, 1 Mar 2002

#from http://code.activestate.com/recipes/117214/
def KnuthMorrisPratt(text, pattern):

    '''Yields all starting positions of copies of the pattern in the text.
Calling conventions are similar to string.find, but its arguments can be
lists or iterators, not just strings, it returns all matches, not just
the first one, and it does not need the whole text in memory at once.
Whenever it yields, it will have read the text exactly up to and including
the match that caused the yield.'''

    # allow indexing into pattern and protect against change during yield
    pattern = list(pattern)

    # build table of shift amounts
    shifts = [1] * (len(pattern) + 1)
    shift = 1
    for pos in range(len(pattern)):
        while shift <= pos and pattern[pos] != pattern[pos-shift]:
            shift += shifts[pos-shift]
        shifts[pos+1] = shift

    # do the actual search
    startPos = 0
    matchLen = 0
    for c in text:
        while matchLen == len(pattern) or \
              matchLen >= 0 and pattern[matchLen] != c:
            startPos += shifts[matchLen]
            matchLen -= shifts[matchLen]
        matchLen += 1
        if matchLen == len(pattern):
            yield startPos

C++程序如下：

#include 
#include 
using namespace std;

//----------------------------
//Returns a vector containing the zero based index of 
//  the start of each match of the string K in S.
//  Matches may overlap
//----------------------------
vector KMP(string S, string K)
{
        vector T(K.size() + 1, -1);
	vector matches;

        if(K.size() == 0)
        {
            matches.push_back(0);
            return matches;
        }
	for(int i = 1; i <= K.size(); i++)
	{
		int pos = T[i - 1];
		while(pos != -1 && K[pos] != K[i - 1]) pos = T[pos];
		T[i] = pos + 1;
	}

	int sp = 0;
	int kp = 0;
	while(sp < S.size())
	{
		while(kp != -1 && (kp == K.size() || K[kp] != S[sp])) kp = T[kp];
		kp++;
		sp++;
		if(kp == K.size()) matches.push_back(sp - K.size());
	}
	
	return matches;
}

C程序如下：

#include
#include
#include
 
void computeLPSArray(char *pat, int M, int *lps);
 
void KMPSearch(char *pat, char *txt)
{
    int M = strlen(pat);
    int N = strlen(txt);
 
    // create lps[] that will hold the longest prefix suffix values for pattern
    int *lps = (int *)malloc(sizeof(int)*M);
    int j  = 0;  // index for pat[]
 
    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pat, M, lps);
 
    int i = 0;  // index for txt[]
    while(i < N)
    {
      if(pat[j] == txt[i])
      {
        j++;
        i++;
      }
 
      if (j == M)
      {
        printf("Found pattern at index %d \n", i-j);
        j = lps[j-1];
      }
 
      // mismatch after j matches
      else if(pat[j] != txt[i])
      {
        // Do not match lps[0..lps[j-1]] characters,
        // they will match anyway
        if(j != 0)
         j = lps[j-1];
        else
         i = i+1;
      }
    }
    free(lps); // to avoid memory leak
}
 
void computeLPSArray(char *pat, int M, int *lps)
{
    int len = 0;  // lenght of the previous longest prefix suffix
    int i;
 
    lps[0] = 0; // lps[0] is always 0
    i = 1;
 
    // the loop calculates lps[i] for i = 1 to M-1
    while(i < M)
    {
       if(pat[i] == pat[len])
       {
         len++;
         lps[i] = len;
         i++;
       }
       else // (pat[i] != pat[len])
       {
         if( len != 0 )
         {
           // This is tricky. Consider the example AAACAAAA and i = 7.
           len = lps[len-1];
 
           // Also, note that we do not increment i here
         }
         else // if (len == 0)
         {
           lps[i] = 0;
           i++;
         }
       }
    }
}
 
// Driver program to test above function
int main()
{
   char *txt = "apurba mandal loves ayoshi loves";
   char *pat = "loves";
   KMPSearch(pat, txt);
   return 0;
}