880 Decoded String at Index

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".

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.

题目大意

计算编码后的字符串的第k个字符，编码规则如例中所示。

思路

直接算出编码后的字符串肯定会爆内存，我们可以首先计算编码后字符串的size，对于本题而言，编码后第k个字符与第k%size的字符是一样的，我们可以倒着遍历字符串，若遇到数字a，则说明数字前的字符串tmp重复a次，则编码后第k个字符与tmp的第k%tmp.length个字符一致，size /= a，即去掉a后的长度为size/a，若遇到字符，则size–。最后直至k=0并且当前字符是字母。

需要注意的是size可能很大，需要用long类型。

代码

class Solution {
    public String decodeAtIndex(String S, int K) {
        long size = 0;
        String s = S;
        for(char i : s.toCharArray()){
            if(Character.isDigit(i)){
                size *= i - '0';
            }
            else{
                size++;
            }
        }
        for(int i = s.length() - 1; i >= 0; i--){
            K %= size;
            if(K == 0 && Character.isLetter(s.charAt(i))){
                return Character.toString(s.charAt(i));
            }
            else if(Character.isLetter(s.charAt(i))){
                size--;
            }
            else{
                size /= s.charAt(i) - '0';
            }
        }
        return null;
    }
}