问题描述：
使用KMP算法快速查找字符串（可支持中文）

PS：
解析KMP的方法虽然多种多样，但是我仍然最喜欢数据结构书中我认为最经典的那种

参考代码：

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int KMPAlgorithm(const wstring& strText, const wstring& strPattern, int *m_pNext, int nStartPos)
{
    int i = nStartPos;
    int j = 0;

    while (i < (int)strText.size() && j < (int)strPattern.size())
    {
        if (j == -1 || strText[i] == strPattern[j])
        {
            ++i;
            ++j;
        }
        else
        {
            j = m_pNext[j];
        }
    }
    if (j == strPattern.size())
        return (i - j);

    return -1;
}

void GetNextVal(const wstring& strPattern, int *m_pNext)
{
    int j = 0;
    int k = -1;
    m_pNext[0] = -1;

    while (j < (int)strPattern.size())
    {
        if (k == -1 || strPattern[j] == strPattern[k])
        {
            ++j;
            ++k;
            if (strPattern[j] != strPattern[k])
                m_pNext[j] = k;
            else
                m_pNext[j] = m_pNext[k];
        }
        else
        {
            k = m_pNext[k];
        }
    }
}

void GetAllSubPos(const wstring& strText, const wstring& strPattern, std::vector<int>& vecans)
{
    size_t szlen = strPattern.size();
    int *m_pNext = new int[strPattern.size() + 1];
    memset(m_pNext, 0, (strPattern.size() + 1) * sizeof(int));

    if (strText.size() == 0 || strPattern.size() == 0)
        return;

    GetNextVal(strPattern, m_pNext);

    vecans.clear();
    int nPos = 0;
    while (1)
    {
        nPos = KMPAlgorithm(strText, strPattern, m_pNext, nPos);
        if (nPos == -1)
            break;

        vecans.push_back(nPos);
        nPos++;
    }

    delete[]m_pNext;
}

int main(int argc, char*argv[])
{
    vector<int>ansvec;
    wstring strText;
    wstring strPattern;
    printf("输入主串: ");
    getline(wcin, strText);
    printf("输入模式串: ");
    getline(wcin, strPattern);
    GetAllSubPos(strText, strPattern, ansvec);
    if (ansvec.size() > 0)
        printf("输入所有查找结果:\n");
    else
        printf("没有查到\n");
    for (int i = 0; i < (int)ansvec.size(); i++)
    {
        wstring strSub = strText.substr(ansvec[i]);
        wcout << strSub << endl;
    }

    return 0;
}

运行结果：