880.Decoded String at Index

Description

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.

题目链接：https://leetcode.com/problems/decoded-string-at-index/description/

Difficulty: medium

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:

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

分析

updating

参考代码

class Solution:
def decodeAtIndex(self, S, K):
    size=0
    for i in S:
        if(i.isdigit()):
            size*=int(i)
        else:
            size+=1
    for index in reversed(S):
        K %= size
        if(K==0 and index.isalpha()):
            return index
        if(index.isdigit()):
            size/=int(index)
        else:
            size-=1
    return