题目例如以下：
Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue",
return "blue is sky the".
Clarification:
- What constitutes a word?
A sequence of non-space characters constitutes a word. - Could the input string contain leading or trailing spaces?
Yes. However, your reversed string should not contain leading or trailing spaces. - How about multiple spaces between two words?
Reduce them to a single space in the reversed string.
题目要求不仅须要把字符串中的单词进行旋转，并且測试用例中还包括头尾含有空格、单词间空格数大于1个之类的字符串。所以要处理好空格，处理完空格的字符串肯定是小于或等于原字符串长度的，考虑到通过移动字符的方式来减小空格耗时长，我的思路是首先计算出去多余的空格后字符串应有的长度，然后又一次定义一个暂时字符串，开辟终于应有长度的空间。然后将原字符串逆序copy到新定义的字符串变量中，保证头尾多余空格已去除，保证单词间仅保留一个空格。然后将每一个单词再进行一次反转，便实现了题目所要求的功能。AC代码例如以下：
class Solution {
public:
void reverseWords(string &s)
{
if(s.empty())
return;
int count = 0;
int index = 0, indexTemp = 0, begin = 0, end = s.length()-1;
while(s[begin] == ' ' && begin<s.length()-1)
begin++;
while(s[end] == ' ' && end>0)
end--;
if(end == 0 && s[end] == ' ')
{
s = "";
return;
}
else if(end == 0)
{
s = s.substr(0,1);
return;
}
index = begin;
while(index <= end)
{
if(s[index] == ' ')
{
count++;
while(s[index] == ' ')
index++;
}
count++;
index++;
}
string temp(count,'\0');
index = end;
indexTemp = 0;
while(index >= begin)
{
if(s[index] == ' ')
{
temp[indexTemp] = s[index];
while(s[index] == ' ')
index--;
indexTemp++;
}
temp[indexTemp] = s[index];
indexTemp++;
index--;
}
indexTemp = 0;
begin = -1;
end = -1;
while(indexTemp < count)
{
if(temp[indexTemp] != ' ' && begin == -1)
{
begin = 0;
}
else if(indexTemp-1 >= 0 && temp[indexTemp-1] == ' '&&temp[indexTemp] != ' ')
{
begin = indexTemp;
}
else if((indexTemp+1 < count && temp[indexTemp+1] == ' ' && temp[indexTemp] != ' ') || ((indexTemp == count-1)&&temp[indexTemp] != ' '))
{
end = indexTemp;
reverse(temp, begin, end);
}
indexTemp++;
}
s = temp;
}
void reverse(string &s, int begin, int end)
{
while(begin < end)
{
char temp = s[begin];
s[begin] = s[end];
s[end] = temp;
begin++;
end--;
}
}
};