880. Decoded String at Index

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# 題目敘述

You are given an encoded string s . To decode the 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 d , the entire current tape is repeatedly written d - 1 more times in total.

Given an integer k , return the kth 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".

# 解題思路

# Solution

class Solution {

    public String decodeAtIndex(String s, int k) {

        long decodedLength = 0; 

        for (char c : s.toCharArray()) {

            if (Character.isDigit(c)) {

                decodedLength *= (c - '0');

            } else {

                decodedLength++;

            }

        }

        for (int i = s.length() - 1; i >= 0; i--) {

            char currChar = s.charAt(i);

            if (Character.isDigit(currChar)) {

                decodedLength /= (currChar - '0');

                k %= decodedLength;

            } else {

                if (k == 0 || decodedLength == k) {

                    return String.valueOf(currChar);

                }

                decodedLength--;

            }

        }

        return "";

    }

}