leetcode之倒转一句话单词

题目：

Given an input string, reverse the string word by word.

For example,
Given s = "the sky is blue",
return "blue is sky the".

click to show 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.
  

  思路：

  这题的思路不用多说，完全是烂大街的题，不就是先以每个单词为单位，进行翻转，再以整句话为单位进行翻转吗？呵呵，太简单了，以为简单你就能做对那可就大错特错了，题干里的的说明还要求去除首尾空格，并压缩多个空格为一个空格，所以会有N多边界条件要考虑，我竟然编译了N次才通过leetcode OJ，作为一个正式工作一年多的程序员，实在是对不起国家，边界条件的确是考察一个人思维严谨性甚至记忆力的试金石！

  代码：
  

  class Solution {
  public:
  	Solution(){}
  	void reverse(std::string &s, int f, int e)
  	{
  		while (e>f)
  		{
  			char tmp = s.at(f);
  			s.at(f) = s.at(e);
  			s.at(e) = tmp;
  			f++;
  			e--;
  		}
  	}
  	void squeeze(std::string &s)
  	{
  		int i = 0;
  		std::string re = "";
  		char prev = NULL;
  		bool flag = false;
  		while (i < s.size())
  		{
  			if (s.at(i)!=' ')
  			{
  				if (flag&&prev == ' ')
  				{
  					re.push_back(' ');
  				}
  				re.push_back(s.at(i));
  			}
  			if (s.at(i) == ' '&&prev !=NULL &&prev!= ' ')
  			{
  				flag = true;
  			}
  			prev = s.at(i);
  			i++;
  		}
  		s = re;
  	}
  	void reverseWords(std::string &s) {
  		int i = 0;
  		int f=0, e=0;
  		char prev = ' ';
  		while (i