604. Design Compressed String Iterator
Difficulty: Easy
Topics: Design
Similar Questions:
Problem:
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 ' ' </pre> </p>
Solutions:
class StringIterator {
public:
StringIterator(string compressedString) {
str = compressedString;
index = 0;
count = 0;
}
char next() {
if (!hasNext()) return ' ';
if (count > 0) {
--count;
return c;
}
c = str[index++];
count = 0;
while (index < str.length() && isdigit(str[index])) {
count = 10 * count + (str[index++] - '0');
}
--count;
return c;
}
bool hasNext() {
return count > 0 || index < str.length();
}
private:
string str;
int index;
char c;
int count;
};
/**
* Your StringIterator object will be instantiated and called as such:
* StringIterator* obj = new StringIterator(compressedString);
* char param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/