KMP prefix table

问题:

I am reading about KMP for string matching.
It needs a preprocessing of the pattern by building a prefix table.
For example for the string ababaca the prefix table is: P = [0, 0, 1, 2, 3, 0, 1]
But I am not clear on what does the numbers show. I read that it helps to find matches of the pattern when it shifts but I can not connect this info with the numbers in the table.

回答1:

Every number belongs to corresponding prefix ("a", "ab", "aba", ...) and for each prefix it represents length of longest suffix of this string that matches prefix. We do not count whole string as suffix or prefix here, it is called self-suffix and self-prefix (at least in Russian, not sure about English terms).

So we have string "ababaca". Let's look at it. KMP computes Prefix Function for every non-empty prefix. Let's define s[i] as the string, p[i] as the Prefix function. prefix and suffix may overlap.

+---+----------+-------+------------------------+
| i |  s[0:i]  | p[i]  | Matching Prefix/Suffix |
+---+----------+-------+------------------------+
| 0 | a        |     0 |                        |
| 1 | ab       |     0 |                        |
| 2 | aba      |     1 | a                      |
| 3 | abab     |     2 | ab                     |
| 4 | ababa    |     3 | aba                    |
| 5 | ababac   |     0 |                        |
| 6 | ababaca  |     1 | a                      |
|   |          |       |                        |
+---+----------+-------+------------------------+

Simple C++ code that computes Prefix function of string S:

vector<int> prefixFunction(string s) {
    vector<int> p(s.size());
    int j = 0;
    for (int i = 1; i < (int)s.size(); i++) {
        while (j > 0 && s[j] != s[i])
            j = p[j-1];

        if (s[j] == s[i])
            j++;
        p[i] = j;
    }   
    return p;
}

回答2:

This code may not be the shortest, but easy to understand flow of code. Simple Java Code for calculating prefix-Array-

    String pattern = "ababaca";
    int i = 1, j = 0;
    int[] prefixArray = new int[pattern.length];
    while (i < pattern.length) {

        while (pattern.charAt(i) != pattern.charAt(j) && j > 0) {
            j = prefixArray[j - 1];

        }
        if (pattern.charAt(i) == pattern.charAt(j)) {
            prefixArray[i] = j + 1;
            i++;
            j++;

        } else {
            prefixArray[i] = j;
            i++;
        }
    }

    for (int k = 0; k < prefixArray.length; ++k) {
        System.out.println(prefixArray[k]);
    }

It produces the required output-

0 0 1 2 3 0 1