leetcode_Count and Say

Problem description:

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.

My solution(python):

class Solution(object):
    def countAndSay(self, n):
        i = 4
        s = str(1)
        slist = ['1', '11', '21']
        for i in range(i, n+1):
            scount = 0 # 记录i+1中字符串的每个位数
            count = 0 # 记录字符串中每组相同数字的个数
            s0 = slist[i-2] # s0代表之前的字符串
            s = '' # s 代表生成新的kong字符串
            k = 0
            for k in range(k, len(s0)):
                if k == len(s0)-1:
                    count += 0
                    s = s + str(count + 1) + s0[k]
                    slist.append(s)
                elif k < len(s0)-1 and s0[k] == s0[k+1]:
                    count += 1
                else:
                    s = s + str(count + 1) + s0[k]
                    count = 0
        return slist[n-1]
        """
        :type n: int
        :rtype: str
        """

I used two layers of circulation.