难度：简单

知识点：字符串

地址：https://leetcode-cn.com/problems/reverse-words-in-a-string-iii

给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。

示例 1:

输入: "Let's take LeetCode contest"
输出: "s'teL ekat edoCteeL tsetnoc" 
注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。

思路

from typing import *
class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        i, j = 0, len(s) - 1
        while i < j:
            if s[i] != s[j]:
                t = s[i]
                s[i] = s[j]
                s[j] = t
            i += 1
            j -= 1
    def reverseWords(self, s: str) -> str:
        words = s.split(" ")
        new_words = []
        for w in words:
            ws = w.split("")
            self.reverseString(ws)
            new_words.append(''.join(ws))
        return ' '.join(new_words)
            

s = Solution()
s.reverseWords("Let's take LeetCode contest") == "s'teL ekat edoCteeL tsetnoc"

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-2-b08e0df35061> in <module>
     24 
     25 s = Solution()
---> 26 s.reverseWords("Let's take LeetCode contest") == "s'teL ekat edoCteeL tsetnoc"

<ipython-input-2-b08e0df35061> in reverseWords(self, s)
     17         new_words = []
     18         for w in words:
---> 19             ws = w.split("")
     20             self.reverseString(ws)
     21             new_words.append(''.join(ws))

ValueError: empty separator