884. Decoded String at Index

An encoded string S is given.  To find and write the decoded string to a tape, the encoded string is read one character at a time and the following steps are taken:

• If the character read is a letter, that letter is written onto the tape.
• If the character read is a digit (say d), the entire current tape is repeatedly written d-1 more times in total.

Now for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the decoded string.

Example 1:

Input: S = "leet2code3", K = 10
Output: "o"
Explanation: 
The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10th letter in the string is "o".

Example 2:

Input: S = "ha22", K = 5
Output: "h"
Explanation: 
The decoded string is "hahahaha".  The 5th letter is "h".

Example 3:

Input: S = "a2345678999999999999999", K = 1
Output: "a"
Explanation: 
The decoded string is "a" repeated 8301530446056247680 times.  The 1st letter is "a".

Note:

1. 2 <= S.length <= 100
2. S will only contain lowercase letters and digits 2 through 9.
3. S starts with a letter.
4. 1 <= K <= 10^9
5. The decoded string is guaranteed to have less than 2^63 letters.

处理到不好处理的时候，就递归一下

class Solution:
    def decodeAtIndex(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        s=set([str(i) for i in range(2,10)])
        i=cnt=0
        while i<len(S):
            j=i
            while j<len(S) and S[j] not in s:
                j+=1
                cnt+=1
                if cnt==K: return S[j-1]
            
            while j<len(S) and S[j] in s:
                if cnt*int(S[j])>=K:
                    left = K%cnt
                    if left==0: left=cnt
                    return self.decodeAtIndex(S[:j], left)
                
                cnt = cnt*int(S[j])
                j+=1
            i=j
            
        return None