c++字串的包含問題，字串包含問題

c++字串的包含問題，字串包含問題

//暴力輪詢法#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，比較快的了,比那些用數組來儲存以及用hash表來儲存更好。#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;}

//還是用數組寫一下吧,hash在這裡沒有什麼意思,建議大家學習第二種方法。#include <iostream>using namespace std;static int b[255];//由於b儲存在全域資料區，會自己初始化。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;}