题目描述：
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 ' '
题目大意：
设计和实现压缩字符串的迭代器，支持函数：next和hasNext。
压缩字符串以字母+出现次数的格式给出。
解题思路：
将压缩字符串compressedString拆分成字母数组chars和出现次数数组times。
记原始字符串大小为size
变量idx记录chars的当前下标，cnt记录当前已经遍历过的字符个数
维护idx和cnt即可。
Java代码：
public class StringIterator {
private ArrayList<Character> chars = new ArrayList<>();
private ArrayList<Long> times = new ArrayList<>();
private long size, cnt;
private int idx;
public StringIterator(String compressedString) {
StringBuilder s = new StringBuilder();
for (char c : (compressedString + "#").toCharArray()) {
if (c >= '0' && c <= '9') {
s.append(c);
continue;
}
if (s.length() > 0) {
size += Integer.parseInt(s.toString());
times.add(this.size);
s = new StringBuilder();
}
if (c != '#') chars.add(c);
}
}
public char next() {
if (!hasNext()) return ' ';
if (times.get(idx) < ++cnt) idx++;
return chars.get(idx);
}
public boolean hasNext() {
return cnt < size;
}
}
/**
* 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();
*/
本文链接：http://bookshadow.com/weblog/2017/06/11/leetcode-design-compressed-string-iterator/
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