# # 【LeetCode】880. Decoded String at Index 解题报告（Python）

## # 题目描述：

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.

## # 解题方法

``````class Solution(object):
def decodeAtIndex(self, S, K):
"""
:type S: str
:type K: int
:rtype: str
"""
size = 0
for c in S:
if c.isdigit():
size *= int(c)
else:
size += 1

for c in reversed(S):
K %= size
if K == 0 and c.isalpha():
return c
if c.isdigit():
size /= int(c)
else:
size -= 1
``````

## # 日期

2018 年 8 月 23 日 ———— 疲惫说明在逆流而上