字符串匹配
void NaiveStringMatching(char *t,char *p)
{
int n=Length(t);
int m=Length(p);
if(n
{
printf(\"Error:The P is longer than T!\\n\");
return;
}
bool find=true;
printf(\"The string T is %s\\n\",t);
printf(\"The string P is %s\\n\",p);
for(int s=0;s
{
if(t[s+i]!=p[i])
{
find=false;
break;
}
}
if(find)
printf(\"Pattern occurs with shift:%d\\n\",s+1);
}
}
int main()
{
char t[]=\"abcdebcg\";
char p[]=\"bcdebcg\";
NaiveStringMatching(t,p);
return 0;
}。
答：strcat(t, s+ i ); s[i]= 0; strcat(s, t );详情>>
