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 ‘ ‘

解法1:

思路比较好想。用一个count记录当前字符还剩下的个数,另一个用pos记录在compressedString里面扫描到的位置。
要注意的是在判断hasNext的时候要check count!=0
C++

1

Java

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public class StringIterator {
int pos = 0;
char prev = ' ';
int count = 0;
String data = null;
public StringIterator(String compressedString) {
this.data = compressedString;
}
public char next() {
if (!hasNext()) {
return ' ';
}
if (count == 0) {
prev = data.charAt(pos++);
int tempCount = 0;
while (pos < data.length() && Character.isDigit(data.charAt(pos))) {
tempCount = tempCount * 10 + data.charAt(pos) - '0';
pos++;
}
count = tempCount;
}
count--;
return prev;
}
public boolean hasNext() {
return pos < data.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();
*/