Rearrange spaces between words in leetcode simple questions

subject

Give you a string text , The string consists of several words surrounded by spaces . Each word consists of one or more lowercase English letters , And there is at least one space between two words . Topic test cases guarantee text Contain at least one word .
Please rearrange the spaces , Make the number of spaces between each pair of adjacent words equal , And try to Maximize The number . If you can't redistribute all spaces equally , please Place extra spaces at the end of the string , This also means that the returned string should be the same as the original text The length of the string is equal .
return Rearrange the string after the spaces .
Example 1：
Input ：text = " this is a sentence "
Output ：“this is a sentence”
explain ： All in all 9 A space and 4 Word . Can be 9 Spaces are evenly distributed between adjacent words , The number of spaces between adjacent words is ：9 / (4-1) = 3 individual .
Example 2：
Input ：text = " practice makes perfect"
Output ：“practice makes perfect "
explain ： All in all 7 A space and 3 Word .7 / (3-1) = 3 A space plus 1 An extra space . Extra spaces need to be placed at the end of the string .
Example 3：
Input ：text = “hello world”
Output ：“hello world”
Example 4：
Input ：text = " walks udp package into bar a”
Output ："walks udp package into bar a "
Example 5：
Input ：text = “a”
Output ：“a”
Tips ：
1 <= text.length <= 100
text Consists of lowercase letters and ’ ’ form
text Contains at least one word
source ： Power button （LeetCode）

Their thinking

   Find the words in the sentence separately , Then calculate the number of spaces and distribute the spaces to each word according to the requirements of the topic .

class Solution:
    def reorderSpaces(self, text: str) -> str:
        words=re.findall(r'\S+',text)
        if len(words)>1:
            a,b=divmod(text.count(' '),len(words)-1)
            return (' '*a).join(words)+b*' '
        else:
            return words[0]+(len(text)-len(words[0]))*' '