The simple problem of leetcode is to split a string into several groups of length K

subject

character string s It can be divided into several lengths according to the following steps k Group ：
The first group consists of the first... In the string k Characters make up , The second group consists of the following k It's a string , And so on . Every character can be One of them Part of a group .
For the last group , If the remaining characters of the string Insufficient k individual , Characters required fill To complete this set of characters .
Be careful , Remove the last group of characters fill（ If it exists ） And connect all groups in order , The resulting string should be s .
Give you a string s , And the length of each group k And a character for filling fill , After processing according to the above steps , Returns an array of strings , The array represents s After grouping Composition of each group .
Example 1：
Input ：s = “abcdefghi”, k = 3, fill = “x”
Output ：[“abc”,“def”,“ghi”]
explain ：
front 3 The characters are “abc” , Form the first group .
Next 3 The characters are “def” , Form a second group .
Last 3 The characters are “ghi” , Form a third group .
Because all groups can be completely filled with characters in the string , So you don't need to use padding characters .
therefore , formation 3 Group , Namely “abc”、“def” and “ghi” .
Example 2：
Input ：s = “abcdefghij”, k = 3, fill = “x”
Output ：[“abc”,“def”,“ghi”,“jxx”]
explain ：
Similar to the previous example , Form the first three groups “abc”、“def” and “ghi” .
For the last group , There are only characters left in the string ‘j’ It can be used . To complete this group , Use padding characters ‘x’ two .
therefore , formation 4 Group , Namely “abc”、“def”、“ghi” and “jxx” .
Tips ：
1 <= s.length <= 100
s It's only made up of lowercase letters
1 <= k <= 100
fill It's a small English letter
source ： Power button （LeetCode）

Their thinking

   You can first fill the string with a length of k Multiple , Then three three separate .

class Solution:
    def divideString(self, s: str, k: int, fill: str) -> List[str]:
        return re.findall('\w{'+str(k)+'}',s if len(s)%k==0 else s+fill*(k-len(s)%k))