151. Reverse Words in a String

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本文介绍了一种算法，用于逐词反转输入字符串中的单词，并确保输出中没有多余的空格。通过两个循环实现，第一个循环用于跳过初始空格，第二个循环在遇到空格前将字符插入临时缓冲区的开头。

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

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

Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.

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.

解法简述：

逐个取元素，第一个循环排除空格，第二个循环在遇到空格或null前将字幕出入缓存单元的第一个位置，这样在遇到空格时为一个正序的单词；

再设定一个缓存string 每获得一个单词就拼接起来末尾加空格；

最后拼接完后，因为排除空格的循环设在第一位，导致最后若存在多个空格时会存入多个空格，因此多设一层循环排除末尾多出空格。

代码如下：

class Solution {
public:
    void reverseWords(string &s) {
        string rs,temp;
        int i = s.length()-1;
        while(s[i] != NULL){
        	while(s[i] == ' ' && s[i] != NULL) i--;
        	while(s[i] != ' ' && s[i] != NULL){
        		temp.insert(0,1,s[i]);
        		i--;
			}		
			rs = rs + temp + " ";
			temp = "";
			i--;
		}
		rs = rs + temp;
		s = rs;
		i = s.length()-1;
		while(s[i] == ' '){
			s.erase(i ,1);
			i--;
		}
    }
};