0880. Decoded String at Index

# 880. Decoded String at Index#

## 题目 #

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 #

• 如果所读的字符是字母，则将该字母写在磁带上。
• 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。

## 代码 #

``````
package leetcode

func isLetter(char byte) bool {
if char >= 'a' && char <= 'z' {
return true
}
return false
}

func decodeAtIndex(S string, K int) string {
length := 0
for i := 0; i < len(S); i++ {
if isLetter(S[i]) {
length++
if length == K {
return string(S[i])
}
} else {
if length*int(S[i]-'0') >= K {
if K%length != 0 {
return decodeAtIndex(S[:i], K%length)
}
return decodeAtIndex(S[:i], length)
}
length *= int(S[i] - '0')
}
}
return ""
}

`````` Apr 8, 2023 Edit this page