LeetCode 604. Design Compressed String Iterator

原题链接在这里：https://leetcode.com/problems/design-compressed-string-iterator/description/

题目：

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 ' '

题解：

用 index 标记当前s位置. 用count计数之前char出现的次数. 当count--到0时继续向后移动 index.

Time Complexity: hasNext(), O(1). next(), O(n). n= compressedString.length().

Space: O(1).

AC Java:

class StringIterator {
    String s;
    int index;
    char cur;
    int count;
    
    public StringIterator(String compressedString) {
        s = compressedString;
        index = 0;
        count = 0;
    }
    
    public char next() {
        if(count != 0){
            count--;
            return cur;
        }
        
        if(index >= s.length()){
            return ' '; 
        }
        
        cur = s.charAt(index++);
        int endIndex = index;
        while(endIndex<s.length() && Character.isDigit(s.charAt(endIndex))){
            endIndex++;
        }
        
        count = Integer.valueOf(s.substring(index, endIndex));
        index = endIndex;
        count--;
        return cur;
    }
    
    public boolean hasNext() {
        return index != s.length() || count != 0;
    }
}

/**
 * Your StringIterator object will be instantiated and called as such:
 * StringIterator obj = new StringIterator(compressedString);
 * char param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

类似String Compression.