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 <iostream> #include <vector> 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<int> KMP(string S, string K) { vector<int> T(K.size() + 1, -1); vector<int> 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<stdio.h> #include<string.h> #include<stdlib.h> 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 ", 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; }