c++字符串的包含问题

//暴力轮询法
#include <iostream>
using namespace std;
int Grial(char *str1, char *str2)
{
    if (str2 == NULL)return 1;
    char *p1 = str1;
    char *p2 = str2;
    while (*p2 != ‘\0‘)
    {
        while (*p1 != ‘\0‘)
        {
            if (*p2 == *p1)
                break;
                p1++;
        }
        if (*p1 == ‘\0‘)
            return 0;
        p2++;
        p1 = str1;
    }
    return 1;
}
int main()
{
    char *s1 = "12345678";
    char *s2 = "2349";
    cout<<Grial(s1, s2)<<endl;
    return 0;
}

//bitset解法，比较快的了,比那些用数组来存储以及用hash表来存储更好。
#include <iostream>
#include <bitset>
using namespace std;
int Grial(char *str1, char *str2)
{
    bitset<255> bt;
    char *p = str1;
    while (*p != ‘\0‘)
    {
        if (bt.test(*p) == 0)
            bt.set(*p);
        p++;
    }
    char *q = str2;
    while (*q != ‘\0‘)
    {
        if (bt.test(*q) == 0)
        {
            return 0;
        }
        q++;
    }
    return 1;
}
int main()
{
    char *s1 = "1231423";
    char *s2 = "239";
    cout<<Grial(s1, s2)<<endl;
    return 0;
}

//还是用数组写一下吧,hash在这里没有什么意思,建议大家学习第二种方法。

#include <iostream>
using namespace std;
static int b[255];//由于b存储在全局数据区，会自己初始化。
int Grial(char *str1,char *str2)
{
    char *p = str1;
    char *q = str2;
    while (*p != ‘\0‘)
    {
        b[*p] = 1;
        p++;
    }
    while (*q != ‘\0‘)
    {
        if (b[*q] == 0)return 0;
        q++;
    }
    return 1;
}
int main()
{
    char *s1 = "123456";
    char *s2 = "16";
    cout<<Grial(s1, s2)<<endl;
    return 0;
}