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104 lines
2.5 KiB
Markdown
104 lines
2.5 KiB
Markdown
# [880. Decoded String at Index](https://leetcode.com/problems/decoded-string-at-index/)
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## 题目
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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:
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If the character read is a letter, that letter is written onto the tape.
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If the character read is a digit (say d), the entire current tape is repeatedly written d-1 more times in total.
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Now for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the decoded string.
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**Example 1**:
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```
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Input: S = "leet2code3", K = 10
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Output: "o"
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Explanation:
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The decoded string is "leetleetcodeleetleetcodeleetleetcode".
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The 10th letter in the string is "o".
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```
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**Example 2**:
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```
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Input: S = "ha22", K = 5
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Output: "h"
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Explanation:
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The decoded string is "hahahaha". The 5th letter is "h".
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```
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**Example 3**:
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```
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Input: S = "a2345678999999999999999", K = 1
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Output: "a"
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Explanation:
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The decoded string is "a" repeated 8301530446056247680 times. The 1st letter is "a".
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```
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**Note**:
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1. 2 <= S.length <= 100
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2. S will only contain lowercase letters and digits 2 through 9.
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3. S starts with a letter.
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4. 1 <= K <= 10^9
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5. The decoded string is guaranteed to have less than 2^63 letters.
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## 题目大意
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给定一个编码字符串 S。为了找出解码字符串并将其写入磁带，从编码字符串中每次读取一个字符，并采取以下步骤：
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- 如果所读的字符是字母，则将该字母写在磁带上。
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- 如果所读的字符是数字（例如 d），则整个当前磁带总共会被重复写 d-1 次。
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现在，对于给定的编码字符串 S 和索引 K，查找并返回解码字符串中的第 K 个字母。
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## 解题思路
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按照题意，扫描字符串扫到数字的时候，开始重复字符串，这里可以用递归。注意在重复字符串的时候到第 K 个字符的时候就可以返回了，不要等所有字符都扩展完成，这样会超时。d 有可能超大。
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## 代码
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```go
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package leetcode
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func isLetter(char byte) bool {
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if char >= 'a' && char <= 'z' {
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return true
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}
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return false
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}
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func decodeAtIndex(S string, K int) string {
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length := 0
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for i := 0; i < len(S); i++ {
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if isLetter(S[i]) {
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length++
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if length == K {
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return string(S[i])
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}
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} else {
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if length*int(S[i]-'0') >= K {
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if K%length != 0 {
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return decodeAtIndex(S[:i], K%length)
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}
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return decodeAtIndex(S[:i], length)
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}
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length *= int(S[i] - '0')
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}
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}
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return ""
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}
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``` |