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
);


}