字符串查找算法之（一）KMP算法

问题：查找Text中是否含有Pattern字符串，返回Pattern在Text中的位置。

 

#include <string.h>#include <iostream>using namespace std;// init the prefix array. when comparing, if text[i] != pattern[j], // then pattern[prefix[j]] should be next check point of pattern against text[i].//// say y = prefix[j], y stands for the biggest length of the word pattern[0..(y-1)]// that makes (pattern[0..(y-1)] == pattern[(j-y+1)..j]). // In another word, y stand for the next comparation point of pattern.// (y==0) means no such word, the next comparation should shart from pattern[0].// Note: it's possible that 0<y<(i-y)<j and 0<(i-y)<y<j.//// When comparing, if (text[i] != pattern[j]) then j = prefix[j-1], // then compare again with text[i] until (j==0)//void InitPrefix(const char *pattern, int prefix[])    {    int len = strlen(pattern);    prefix[0] = 0;    for (int i = 1; i < len; i++)        {        int k = prefix[i - 1];        while ((k > 0) && (pattern[i] != pattern[k]))            {            k = prefix[k - 1];            }        if (pattern[i] == pattern[k])            {            prefix[i] = k + 1;            }        else            {            prefix[i]=0;            }        }    return;    }int StrStr(const char* text, const char* pattern)    {    if (!text || !pattern || pattern[0] == '/0' || text[0] == '/0')        {        return -1;        }    int lenText = strlen(text);    int lenPattern = strlen(pattern);    if (lenText < lenPattern)         {        return -1;        }    int *prefix = new int[lenPattern + 1];    InitPrefix(pattern, prefix); //the prefix array of pattern    int index = -1; // the index to be returned.    int i = 0; // the compare index in text.    int j = 0; // the compare index in pattern    while ((text[i] != '/0') && ((lenText - i) >= (lenPattern - j)))        {        if (pattern[j] == text[i])            {            //reach the end of pattern, find match! return.            if (pattern[j+1] == '/0')                {                index = i - lenPattern + 1;                break;                }            // compare next char in pattern and text.            j++;            i++;             }        else            {            if (j == 0)                {                // pattern[0] != text[i], then skip to text[i+1].                i++;                }            else                 {                //pattern[j]!= text[i], the next check point should be pattern[prefix[j-1]]                j = prefix[j-1];                }            }        }    delete []prefix;    prefix=0;    return index;    }