MS面试曾出过这样一道题目，将给定字符串按单词翻转，如，"This is Charles" -> "Charles is This"
方法一，需要额外缓冲区
1
#include
<
cstring
>
2
void
ReverseByWords(
char
*
s,
char
seperator)
3
{
4
int
iRevTokenScanner
=
strlen(s)
-
1
;
5
char
*
buffer
=
new
char
[iRevTokenScanner
+
2
]();
6
int
iWritePos
=
0
;
7
int
iWordEnd, iWordReadPos;
8
9
while
(iRevTokenScanner
>=
0
)
10
{
11
if
(s[iRevTokenScanner]
!=
seperator)
12
{
13
iWordEnd
=
iRevTokenScanner;
14
15
while
(iRevTokenScanner
>=
0
&&
s[iRevTokenScanner]
!=
seperator)
16
iRevTokenScanner
--
;
17
18
iWordReadPos
=
iRevTokenScanner
+
1
;
19
20
while
(iWordReadPos
<=
iWordEnd)
21
buffer[iWritePos
++
]
=
s[iWordReadPos
++
];
22
}
23
else
24
{
25
buffer[iWritePos
++
]
=
s[iRevTokenScanner
--
];
26
}
27
}
28
29
buffer[iWritePos]
=
'
\0
'
;
30
strcpy(s, buffer);
31
32
if
(buffer)
33
{
34
delete [] buffer;
35
buffer
=
0
;
36
}
37
}
方法二，个人比较prefer
1void ReverseString(char *s, int start, int end)
2{
3 while (start < end)
4 {
5 if (s[start] != s[end])
6 {
7 s[start] ^= s[end];
8 s[end] ^= s[start];
9 s[start] ^= s[end];
10 }
11
12 start++;
13 end--;
14 }
15}
16
17void ReverseByWords(char *s, int len, char seperator)
18{
19 int start = 0, end = 0;
20
21 ReverseString(s, start, len - 1);
22
23 while (end < len)
24 {
25 if (s[end] != seperator)
26 {
27 start = end;
28
29 while (end < len && s[end] != seperator)
30 end++;
31 end--;
32
33 ReverseString(s, start, end);
34 }
35
36 end++;
37
38 }//while
39}
等有空再回来作注释吧，现在挺忙的，先把code贴出来
posted on 2006-11-16 18:50
Charles 阅读(419)
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