反转字符串的单词

对于一个给定的字符串，我们需要在线性(也就是O(n))的时间里对它做一些变形。首先这个字符串中包含着一些空格，就像"Hello World"一样，然后我们要做的是把着个字符串中由空格隔开的单词反序，同时反转每个字符的大小写。比如"Hello World"变形后就变成了"wORLD hELLO"。


输入描述:

给定一个字符串s以及它的长度n(1≤n≤500)

输出描述:

请返回变形后的字符串。题目保证给定的字符串均由大小写字母和空格构成。

输入例子:

"This is a sample",16

输出例子:

"SAMPLE A IS tHIS"

#include <string>
#include <iostream>
using namespace std;
string reverse(string s, int begin, int end)
{

if (s == "" || end <= 0)
return s;
for (int i = begin; i<(begin+end+1) / 2; ++i)
{
char temp;
temp = s[i];
s[i] = s[end - i + begin];
s[end - i + begin] = temp;
}

return s;
}

int main(void) {
string s = "";
char kong = ' ';
int n = 0;
while (getline(cin, s)) {
n = s.size();
cout << "n=" << n << endl;
int begin = 0;
int end = n - 1;

s = reverse(s, begin, end);
//cout << s<<endl;
end = 0;
while (end < n - 1)
{
end++;
if (s[end] != kong)
continue;
else
{
end--;
s = reverse(s, begin, end);

end += 2;
begin = end;

while (s[end] == kong) {
begin++;
end++;
}
}
}

s = reverse(s, begin, end);
//	cout << s << endl;
for (int i = 0; i < s.size(); i++) {
if (s[i] >= 'a'&& s[i] <= 'z')
s[i] -= 'a' - 'A';
else if (s[i] >= 'A'&& s[i] <= 'Z')
s[i] += 'a' - 'A';
else {}
}
cout << s << endl;
}
}