Given an input string s, reverse the order of the words.
A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.
Return a string of the words in reverse order concatenated by a single space.
\n\nNote that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.
\n
Example 1:
\n\n\nInput: s = "the sky is blue"\nOutput: "blue is sky the"\n\n\n
Example 2:
\n\n\nInput: s = " hello world "\nOutput: "world hello"\nExplanation: Your reversed string should not contain leading or trailing spaces.\n\n\n
Example 3:
\n\n\nInput: s = "a good example"\nOutput: "example good a"\nExplanation: You need to reduce multiple spaces between two words to a single space in the reversed string.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 104s contains English letters (upper-case and lower-case), digits, and spaces ' '.s.\n
Follow-up: If the string data type is mutable in your language, can you solve it in-place with O(1) extra space?
给你一个字符串 s ，请你反转字符串中 单词 的顺序。
单词 是由非空格字符组成的字符串。s 中使用至少一个空格将字符串中的 单词 分隔开。
返回 单词 顺序颠倒且 单词 之间用单个空格连接的结果字符串。
\n\n注意：输入字符串 s中可能会存在前导空格、尾随空格或者单词间的多个空格。返回的结果字符串中，单词间应当仅用单个空格分隔，且不包含任何额外的空格。
\n\n
示例 1：
\n\n\n输入：s = \"\n\nthe sky is blue\"\n输出：\"blue is sky the\"\n
示例 2：
\n\n\n输入：s = \" hello world \"\n输出：\"world hello\"\n解释：反转后的字符串中不能存在前导空格和尾随空格。\n\n\n
示例 3：
\n\n\n输入：s = \"a good example\"\n输出：\"example good a\"\n解释：如果两个单词间有多余的空格，反转后的字符串需要将单词间的空格减少到仅有一个。\n\n\n
\n\n
提示：
\n\n1 <= s.length <= 104s 包含英文大小写字母、数字和空格 ' 's 中 至少存在一个 单词\n\n
进阶：如果字符串在你使用的编程语言中是一种可变数据类型，请尝试使用 O(1) 额外空间复杂度的 原地 解法。
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