#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 patternint *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 matcheselse if (i < N && pat[j] != txt[i]){// Do not match lps[0..lps[j-1]] characters,// they will match anywayif (j != 0)j = lps[j-1];elsei = i+1;}}free(lps); // to avoid memory leak
}void computeLPSArray(char *pat, int M, int *lps)
{int len = 0;  // length of the previous longest prefix suffixint i;lps[0] = 0; // lps[0] is always 0i = 1;// the loop calculates lps[i] for i = 1 to M-1while (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 = "ABABDABACDABABCABAB";char *pat = "ABABCABAB";KMPSearch(pat, txt);return 0;
}

现在Python语言也是比较流行的，再用Python实现一遍代码如下所示：

# Python program for KMP Algorithm
def KMPSearch(pat, txt):M = len(pat)N = len(txt)# create lps[] that will hold the longest prefix suffix # values for patternlps = [0]*Mj = 0 # index for pat[]# Preprocess the pattern (calculate lps[] array)computeLPSArray(pat, M, lps)i = 0 # index for txt[]while i < N:if pat[j] == txt[i]:i += 1j += 1if j == M:print "Found pattern at index " + str(i-j)j = lps[j-1]# mismatch after j matcheselif i < N and pat[j] != txt[i]:# Do not match lps[0..lps[j-1]] characters,# they will match anywayif j != 0:j = lps[j-1]else:i += 1def computeLPSArray(pat, M, lps):len = 0 # length of the previous longest prefix suffixlps[0] # lps[0] is always 0i = 1# the loop calculates lps[i] for i = 1 to M-1while i < M:if pat[i]==pat[len]:len += 1lps[i] = leni += 1else:# This is tricky. Consider the example.# AAACAAAA and i = 7. The idea is similar # to search step.if len != 0:len = lps[len-1]# Also, note that we do not increment i hereelse:lps[i] = 0i += 1txt = "ABABDABACDABABCABAB"
pat = "ABABCABAB"
KMPSearch(pat, txt)

当然Java语言也是必不可少的，Java编写的代码如下所示：

class KMP_String_Matching
{void KMPSearch(String pat, String txt){int M = pat.length();int N = txt.length();// create lps[] that will hold the longest// prefix suffix values for patternint lps[] = new 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.charAt(j) == txt.charAt(i)){j++;i++;}if (j == M){System.out.println("Found pattern "+"at index " + (i-j));j = lps[j-1];}// mismatch after j matcheselse if (i < N && pat.charAt(j) != txt.charAt(i)){// Do not match lps[0..lps[j-1]] characters,// they will match anywayif (j != 0)j = lps[j-1];elsei = i+1;}}}void computeLPSArray(String pat, int M, int lps[]){// length of the previous longest prefix suffixint len = 0;int i = 1;lps[0] = 0;  // lps[0] is always 0// the loop calculates lps[i] for i = 1 to M-1while (i < M){if (pat.charAt(i) == pat.charAt(len)){len++;lps[i] = len;i++;}else  // (pat[i] != pat[len]){// This is tricky. Consider the example.// AAACAAAA and i = 7. The idea is similar // to search step.if (len != 0){len = lps[len-1];// Also, note that we do not increment// i here}else  // if (len == 0){lps[i] = len;i++;}}}}// Driver program to test above functionpublic static void main(String args[]){String txt = "ABABDABACDABABCABAB";String pat = "ABABCABAB";new KMP_String_Matching().KMPSearch(pat,txt);}
}

算法也是检验开发者处理问题的能力，在业余时间没事的时候，可以编写一下就当放松了。。。。。。。