判断一个字符串是否由另一个字符串旋转而成

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
);



}