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

标签（空格分隔）： LeetCode

作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/decoded-string-at-index/description/

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.

题目大意

从左到右是一个编码了的字符串，如果某个位置是字符，那么拼接到前面的字符串上去；如果某个位置是数字，那么将左边所有的字符串*这个数字的倍数，变成新的字符串。

解题方法

如果模拟题目中的操作进行解码，空间占用过大，一定会通不过。而且，题目只要求了求指定位置的字符，并没有要求所有的字符，所以全部解码没有必要。

参考了官方的解答。思路如下：

比如，对于一个解码了的字符串，appleappleappleappleappleapple ，并且要求的索引K=24的话，那么结果和K=4是一样的。因为单词apple的size=5，重复了6次。所以第K个索引和第K%size个索引是一样的。

所以我们使用反向的计算，保持追踪解码字符串的size，如果解码字符串等于一个word重复了d次的时候，我们可以把K变化为K % (word.length).

算法：

首先，找出解码字符串的长度。然后，我们反向查找，保持追踪size，也就是对编码字符串S[0], S[1], ..., S[i]解码后的长度。

如果找到的是一个数字S[i]，那么意味着解码字符串S[0], S[1], ..., S[i-1]的长度应该是size / Integer(S[i])；否则应该是size-1.

代码如下：

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

参考资料：https://leetcode.com/problems/decoded-string-at-index/solution/

日期

2018 年 8 月 23 日 ———— 疲惫说明在逆流而上