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。