1.strstr
int strstr(char *string, char *substring)
{
if (string == NULL || substring == NULL)
return -1;
int lenstr = strlen(string);
int lensub = strlen(substring);
if (lenstr < lensub)
return -1;
int len = lenstr - lensub;
for (int i = 0; i <= len; i++) //复杂度为O(m*n)
{
for (int j = 0; j < lensub; j++)
{
if (string[i+j] != substring[j])
break;
}
if (j == lensub)
return i + 1;
}
return -1;
}
2.strcpy
//得10分,基本上所有的情况,都考虑到了
//如果有考虑到源目所指区域有重叠的情况,加1分!
char * strcpy( char *strDest, const char *strSrc )
{
if(strDest == strSrc) { return strDest; }
assert( (strDest != NULL) && (strSrc != NULL) );
char *address = strDest;
while( (*strDest++ = * strSrc++) != '/0' );
return address;
}
3.