C拼图 [英] C puzzle
问题描述
给定一个字符串s1和一个字符串s2，写一个片段来说明s2
是否仅仅使用了s1的
轮换一次调用strstr例程？
（例如，给定s1 = ABCD和s2 = CDAB，返回true，
给定s1 = ABCD，s2 = ACBD，返回false）
deepak说：
>
给定一个字符串s1和一个字符串s2，写一个片段，说明s2
是否仅使用一次调用strstr的s1转换为
常规？
肯定那是'你的工作？
-
Richard Heathfield
Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上述域名中， - www。
您好，
给定一个字符串s1和一个字符串s2，写一个片段，说明s2
是否仅使用一次调用strstr例程的s1旋转
？
（例如，给定s1 = ABCD和s2 = CDAB，返回true，
给定s1 = ABCD，s2 = ACBD，返回false）
这并不是很复杂。一个可能的算法包括：
1）找到n，这样s2 [n] == s1 [0]
2）构建一个临时字符串y这是s2旋转了n个位置。
3）比较s1和y。
干杯，
Loic。
lo ****** @ gmx。 net 写道：
你好，
给定一个字符串s1和一个字符串s2，只用一次调用strstr例程，写一个片段来说明是否s2
是s1的旋转？
（例如给定s1 = ABCD且s2 = CDAB，返回true，
给定s1 = ABCD，s2 = ACBD，返回false）
并不是很复杂。一个可能的算法包括：
1）找到n，这样s2 [n] == s1 [0]
2）构建一个临时字符串y这是s2旋转n个位置。
3）比较s1和y。
如果
$ b这可能不起作用$ b s1 = ABCADE和s2 = CADEAB
s2 [n] = A，其中n = 1.临时字符串y = ADEABC。
s1比较不等于y。
我的意思是说你的算法会失败，如果字符串s2中有多个s1 [0]
。
Given a string s1 and a string s2, write a snippet to say whether s2
is a
rotation of s1 using only one call to strstr routine?
(eg given s1 = ABCD and s2 = CDAB, return true,
given s1 = ABCD, and s2 = ACBD , return false)
deepak said:
>
Given a string s1 and a string s2, write a snippet to say whether s2
is a
rotation of s1 using only one call to strstr routine?Surely that''s your job?
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Hello,
Given a string s1 and a string s2, write a snippet to say whether s2
is a
rotation of s1 using only one call to strstr routine?
(eg given s1 = ABCD and s2 = CDAB, return true,
given s1 = ABCD, and s2 = ACBD , return false)This is not really complicated. A possible algorithm consists to:
1) Find n, such that s2[n] == s1[0]
2) Build a temporary string y which is s2 rotated of n positions.
3) Compare s1 and y.
Cheers,
Loic.
lo******@gmx.net wrote:Hello,
Given a string s1 and a string s2, write a snippet to say whether s2
is a
rotation of s1 using only one call to strstr routine?
(eg given s1 = ABCD and s2 = CDAB, return true,
given s1 = ABCD, and s2 = ACBD , return false)
This is not really complicated. A possible algorithm consists to:
1) Find n, such that s2[n] == s1[0]
2) Build a temporary string y which is s2 rotated of n positions.
3) Compare s1 and y.
This may not work if
s1 = ABCADE and s2 = CADEAB
s2[n] = A, where n = 1. temporary string y = ADEABC.
s1 compares not equal to y.
I mean to say your algorithm will fail, if there is more than one s1[0]
in string s2.
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