[leetcode]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 ' '

思路：逐一判断每个字符，遇到是数字的时候就循环前一个字符N次，注意数字有可能是大于10，即不一定是一位数。

代码如下：

package com.billkang;

public class StringIterator {

	private String compressedString;
	int c = 0;
	private int pos = 0;
	private char ch;

	public StringIterator(String compressedString) {
		this.compressedString = compressedString;
	}

	public char next() {
		if (c > 0) {
			c--;
			return ch;
		}

		if (pos >= compressedString.length()) {
			return ' ';
		}

		ch = compressedString.charAt(pos++);
		while (pos < compressedString.length()
				&& isDigit(compressedString.charAt(pos))) {
			c = c * 10 + compressedString.charAt(pos++) - '0';
		}
		c--;
		return ch;
	}

	private boolean isDigit(char charAt) {
		return Character.isDigit(charAt);
	}

	public boolean hasNext() {
		return c > 0 || pos < compressedString.length();
	}
}