Given s = “the sky is blue”,
* return “blue is sky the”.
這種問題一般有三種解法：
1，One simple approach is a two-pass solution: First pass to split the string by spaces into an array of words, then second pass to extract the words in reversed order.
這種方法需要額外空間儲存單詞，而且需要遍歷兩邊。
2，We can do better in one-pass. While iterating the string in reverse order, we keep track of a word’s begin and end position. When we are at the beginning of a word, we append it.
需要額外空間，從後面遍歷，只需要遍歷一次。（當然如果用C，需要先遍歷一次獲取尾部）
3，in-place 方法：需要O(1)的額外空間,多次翻轉的方法；（如果單詞之間以前開頭和結尾處有很多空格，處理會麻煩一些，很考驗人）

Reverse Words in a String .II
* Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.
*
* The input string does not contain leading or trailing spaces and the words are always separated by a single space.
*
* For example,
* Given s = “the sky is blue”,
* return “blue is sky the”.
*
* Could you do it in-place without allocating extra space?
如果不考慮空格問題，非常容易利用多次反轉解決。
//先定義reverse函式，翻轉字串

void reverse(char *s,int start, int last){
    while(start<last){
        char temp=s[start];
        s[start++]=s[last];
        s[last--]=temp;
    }
}

void reverseWords_no_space(char *s){
  int j;//len
  for(j=0;s[j]!='\0';j++);//找到尾部
  reverse(s,0,j-1);
  for(int begin=0,i=0;i<=j;i++){
    if(s[i]==' '
 
||s[i]=='\0'){
        reverse(s,begin,i-1);
        begin=i+1;
    }
  }

}

Reverse Words in a String
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.

void reverse(char *s,int start, int last){
    while(start<last){
        char temp=s[start];
        s[start++]=s[last];
        s[last--]=temp;
    }
}

void reverseWords(char *s) {//只有一個單詞，就把兩邊的空格去掉
    int index=0;//先用於記錄長度
    //i為不為空格的第一位
    for(index=0;s[index]!='\0';index++);//找到尾部
    reverse(s,0,index-1);
    for(int i=0,begin=0;i<=index;i++){//反轉單詞
        if(s[i]==' '||s[i]=='\0') continue;
        else {
            begin=i;
            for(;s[i]!=' '&&s[i]!='\0';i++);
            reverse(s,begin,i-1);
            begin=i+1;
         }
    }
    for(int i=0,newi=0,extraspace=true;i<=index;i++){
        if(s[i]!=' '||(s[i]==' '&&!extraspace)){//保證單詞之間以及開頭處不會有多餘的空格
            if(s[i]!='\0') s[newi++]=s[i];
            else {
                if(newi>0&&s[newi-1]==' ') s[--newi]='\0';//處理結尾的空格
                else s[newi]='\0';
                return;
            }
            if(s[i]==' ') extraspace=true;
            else extraspace=false;
        }
    }
}