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.

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.

题目解析：

题目中要删除多余的空格，有一种比较简单的方法，先将s字符串赋给一个临时字符串tmp，将头尾空格略去，两个字间的空格只复制一次。然后对tmp进行常规操作。这样需要两遍遍历。

也可以临时设一个结果字符串，和一个临时的字串。一点一点处理

class Solution {
public:
    void reverseWords(string &s) {
        string tmp;
        for(int i = s.size()-1;i >= 0;i--){
            while(i >= 0 && s[i] == ' ')
                i--;
            if(i<0)
                break;
            if(!tmp.empty())
                tmp.push_back(' ');
            string word;
            while(i>=0 && s[i]!=' ')
                word.push_back(s[i--]);
            reverse(word.begin(),word.end());
            tmp.append(word);
        }
        s = tmp;
    }
};