# 880. Decoded String at Index

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.

``````class Solution {
public:
string decodeAtIndex(string S, int K) {
long i = 0, cnt = 0;
for (; cnt < K; ++i) {
cnt = isdigit(S[i]) ? cnt * (S[i] - '0') : (cnt + 1);
}
while (i--) {
if (isdigit(S[i])) {
cnt /= (S[i] - '0');
K %= cnt;
} else {
if (K % cnt == 0) return string(1, S[i]);
--cnt;
}
}
return "grandyang";
}
};
``````

``````class Solution {
public:
string decodeAtIndex(string S, int K) {
long cnt = 0;
for (int i = 0; i < S.size(); ++i) {
if (isalpha(S[i])) {
if (++cnt == K) return string(1, S[i]);
} else {
if (cnt * (S[i] - '0') >= K) return decodeAtIndex(S.substr(0, i), (K - 1) % cnt + 1);
cnt *= (S[i] - '0');
}
}
return "grandyang";
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/880

https://leetcode.com/problems/decoded-string-at-index/

https://leetcode.com/problems/decoded-string-at-index/discuss/157156/15-lines-clear-code

https://leetcode.com/problems/decoded-string-at-index/discuss/156747/C%2B%2BPython-O(N)-Time-O(1)-Space

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation