1 void computeLPSArray(char *pat, int M, int *lps)
 2 {
 3     // length of the previous longest prefix suffix
 4     int len = 0;
 5
 6     lps[0] = 0; // lps[0] is always 0
 7
 8     // the loop calculates lps[i] for i = 1 to M-1
 9     int i = 1;
10     while (i < M)
11     {
12         if (pat[i] == pat[len])
13         {
14             len++;
15             lps[i] = len;
16             i++;
17         }
18         else // (pat[i] != pat[len])
19         {
20             // This is tricky. Consider the example.
21             // AAACAAAA and i = 7. The idea is similar
22             // to search step.
23             if (len != 0)
24             {
25                 len = lps[len-1];
26
27                 // Also, note that we do not increment
28                 // i here
29             }
30             else // if (len == 0)
31             {
32                 lps[i] = 0;
33                 i++;
34             }
35         }
36     }
37 }

 1 void KMPSearch(char *pat, char *txt)
 2 {
 3     int M = strlen(pat);
 4     int N = strlen(txt);
 5
 6     // create lps[] that will hold the longest prefix suffix
 7     // values for pattern
 8     int lps[M];
 9
10     // Preprocess the pattern (calculate lps[] array)
11     computeLPSArray(pat, M, lps);
12
13     int i = 0;  // index for txt[]
14     int j  = 0;  // index for pat[]
15     while (i < N)
16     {
17         if (pat[j] == txt[i])
18         {
19             j++;
20             i++;
21         }
22
23         if (j == M)
24         {
25             printf("Found pattern at index %d \n", i-j);
26             j = lps[j-1];
27         }
28
29         // mismatch after j matches
30         else if (i < N && pat[j] != txt[i])
31         {
32             // Do not match lps[0..lps[j-1]] characters,
33             // they will match anyway
34             if (j != 0)
35                 j = lps[j-1];
36             else
37                 i = i+1;
38         }
39     }
40 }

 1 void preparation(char *P, int *f) {
 2     int m = strlen(P);
 3     f[0] = f[1] = 0;
 4     for (int i = 1; i < m; i++) {
 5         int j = f[i];
 6         while (j && P[i] != P[j])
 7               j = f[j];
 8         f[i + 1] = (P[i] == P[j])? j+1 : 0;
 9     }
10 }
11 void KMP(char *T, char *P, int *f) {
12     int n = strlen(T), m = strlen(P);
13     preparation(P, f);
14     int j = 0;
15     for (int i = 0; i < n; i++) {
16         while (j && P[i] != T[i]) j = f[j];
17         if (P[j] == T[i]) j++;
18         if (j == m) answer(i - m + 1);
19     }
20 }