604. Design Compressed String Iterator

Design and implement a data structure for a compressed string iterator. It should support the following operations: next and hasNext.
The given compressed string will be in the form of each letter followed by a positive integer representing the number of this letter existing in the original uncompressed string.
next() - if the original string still has uncompressed characters, return the next letter; Otherwise return a white space.
hasNext() - Judge whether there is any letter needs to be uncompressed.
Note:
Please remember to RESET your class variables declared in StringIterator, as static/class variables are persisted across multiple test cases. Please see here for more details.
Example:
StringIterator iterator = new StringIterator("L1e2t1C1o1d1e1");

iterator.next(); // return 'L'
iterator.next(); // return 'e'
iterator.next(); // return 'e'
iterator.next(); // return 't'
iterator.next(); // return 'C'
iterator.next(); // return 'o'
iterator.next(); // return 'd'
iterator.hasNext(); // return true
iterator.next(); // return 'e'
iterator.hasNext(); // return false
iterator.next(); // return ' '



https://leetcode.com/problems/design-compressed-string-iterator/solution/

 

public class StringIterator {
    
    Queue<int[]> queue = new LinkedList<>();
    
    public StringIterator(String s) {
        int i = 0, n = s.length();
        while (i < n) {
            int j = i+1;
            while (j < n && s.charAt(j) - 'A' < 0) j++;
            queue.add(new int[]{s.charAt(i) - 'A',  Integer.parseInt(s.substring(i+1, j))});
            i = j;
        }
    }
    
    public char next() {
        if (queue.isEmpty()) return ' ';
        int[] top = queue.peek();
        if (--top[1] == 0) queue.poll();
        return (char) ('A' + top[0]);
    }
    
    public boolean hasNext() {
        return !queue.isEmpty();
    }

}