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.

Input: S = "leet2code3", K = 10
Output: "o"
Explanation: 
The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10th letter in the string is "o".

Input: S = "ha22", K = 5
Output: "h"
Explanation: 
The decoded string is "hahahaha".  The 5th letter is "h".

Input: S = "a2345678999999999999999", K = 1
Output: "a"
Explanation: 
The decoded string is "a" repeated 8301530446056247680 times.  The 1st letter is "a".

Constraints:

• 2 <= S.length <= 100
• S will only contain lowercase letters and digits 2 through 9.
• S starts with a letter.
• 1 <= K <= 10^9
• It‘s guaranteed that K is less than or equal to the length of the decoded string.
• The decoded string is guaranteed to have less than 2^63 letters.

class Solution {
    public String decodeAtIndex(String S, int K) {
        long curlength = 0;
        int n = S.length();
        
        for(int i = 0; i < n; i++) {
            char c = S.charAt(i);
            if(Character.isDigit(c)) {
                curlength *= c - ‘0‘;
            }
            else curlength += 1;
        }
        
        for(int i = n - 1; i >= 0; i--) {
            char c = S.charAt(i);
            if(Character.isDigit(c)) {
                curlength /= c - ‘0‘;
                K %= curlength;
            }
            else {
                if(K == 0 || K == curlength) return c + "";
                else curlength--;
            }
        }
        return "";
    }
}

https://leetcode.com/problems/decoded-string-at-index/discuss/157390/Logical-Thinking-with-Clear-Code

先把总长度求出来，然后从后往前减小长度，直到当前长度==k为止

k==0是当出现a2这种情况，完事后k就==0了

880. Decoded String at Index

原文：https://www.cnblogs.com/wentiliangkaihua/p/14166234.html