if s1 = "stackoverflow"
then the following are some of its rotated versions:
"tackoverflows"
"ackoverflowst"
"overflowstack"
where as "stackoverflwo"
is not
a rotated version.
通常的做法
algorithm checkRotation ( string s1 , string s2 )
if ( len ( s1 ) != len ( s2 ))
return false
if ( substring ( s2 , concat ( s1 , s1 ))
return true
return false
end
一个基于KMP的解决方案
bool is_rotation ( const string & str1 , const string & str2 )
{
if ( str1 . size ()!= str2 . size ())
return false ;
vector <size_t> prefixes ( str1 . size (), 0 );
for ( size_t i = 1 , j = 0 ; i < str1 . size (); i ++) {
while ( j > 0 && str1 [ i ]!= str1 [ j ])
j = prefixes [ j - 1 ];
if ( str1 [ i ]== str1 [ j ]) j ++;
prefixes [ i ]= j ;
}
size_t i = 0 , j = 0 ;
for (; i < str2 . size (); i ++) {
while ( j > 0 && str2 [ i ]!= str1 [ j ])
j = prefixes [ j - 1 ];
if ( str2 [ i ]== str1 [ j ]) j ++;
}
for ( i = 0 ; i < str2 . size (); i ++) {
if ( j >= str1 . size ()) return true ;
while ( j > 0 && str2 [ i ]!= str1 [ j ])
j = prefixes [ j - 1 ];
if ( str2 [ i ]== str1 [ j ]) j ++;
}
return false ;
}
c版本
int is_rotation ( char * s1 , char * s2 )
{
char * tmp1 ;
char * tmp2 ;
char * ref2 ;
assert ( s1 && s2 );
if (( s1 == s2 ) || ( strcmp ( s1 , s2 ) == 0 ))
return ( 1 );
if ( strlen ( s1 ) != strlen ( s2 ))
return ( 0 );
while (* s2 )
{
tmp1 = s1 ;
if (( ref2 = strchr ( s2 , * s1 )) == NULL )
return ( 0 );
tmp2 = ref2 ;
while (* tmp1 && (* tmp1 == * tmp2 ))
{
++ tmp1 ;
++ tmp2 ;
if (* tmp2 == '/0' )
tmp2 = s2 ;
}
if (* tmp1 == '/0' )
return ( 1 );
else
++ s2 ;
}
return ( 0 );
}