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 
#include 

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 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;
    }