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

Example 1:

Input: "the sky is blue" Output: "blue is sky the" 

Example 2:

Input: " hello world! " Output: "world! hello" Explanation: Your reversed string should not contain leading or trailing spaces. 

Example 3:

Input: "a good example" Output: "example good a" Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string. 

Note:

• A word is defined as a sequence of non-space characters.
• Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
• You need to reduce multiple spaces between two words to a single space in the reversed string.

Follow up:

For C programmers, try to solve it in-place in O(1) extra space.

给定一个字符串，逐个翻转字符串中的每个单词。

说明：

• 无空格字符构成一个单词。
• 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
• 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。  

进阶：

• 请选用 C 语言的用户尝试使用 O(1) 额外空间复杂度的原地解法。

• 给出一个中间有空格分隔的字符串，要求把这个字符串按照单词的维度前后翻转。
• 依照题意，先把字符串按照空格分隔成每个小单词，然后把单词前后翻转，最后再把每个单词中间添加空格。