604. Design Compressed String Iterator

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Problem

Design and implement a data structure for a compressed string iterator. 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.

Implement the StringIterator class:

• next() Returns the next character if the original string still has uncompressed characters, otherwise returns a white space.
• hasNext() Returns true if there is any letter needs to be uncompressed in the original string, otherwise returns false.

Example 1:

Input
["StringIterator", "next", "next", "next", "next", "next", "next", "hasNext", "next", "hasNext"]
[["L1e2t1C1o1d1e1"], [], [], [], [], [], [], [], [], []]
Output
[null, "L", "e", "e", "t", "C", "o", true, "d", true]

Explanation
StringIterator stringIterator = new StringIterator("L1e2t1C1o1d1e1");
stringIterator.next(); // return "L"
stringIterator.next(); // return "e"
stringIterator.next(); // return "e"
stringIterator.next(); // return "t"
stringIterator.next(); // return "C"
stringIterator.next(); // return "o"
stringIterator.hasNext(); // return True
stringIterator.next(); // return "d"
stringIterator.hasNext(); // return True

Constraints:

• 1 <= compressedString.length <= 1000
• compressedString consists of lower-case an upper-case English letters and digits.
• The number of a single character repetitions in compressedString is in the range $[1, 10^9]$
• At most 100 calls will be made to next and hasNext.

Code

class StringIterator {
    String str;
    char curr;
    int count;
    int index;
  
    public StringIterator(String compressedString) {
        str = compressedString;
        curr = ' ';
        index = 0;
        count = 0;
    }
    
    public char next() {
        if(!hasNext()) return ' ';

        if (count == 0) {
            curr = str.charAt(index++);

            while (index < str.length() && Character.isDigit(str.charAt(index))) {
                count = 10 * count + (str.charAt(index) - '0');
                index++;
            }
        }

        count--;
        return curr;
    }
    
    public boolean hasNext() {
        return index < str.length() || count != 0;
    }
}