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 ' 'class StringIterator {
public:
StringIterator(string compressedString) {
int i = 0, n = compressedString.length();
while (i < n) {
int j = i + 1;
while (j < n && compressedString[j] - 'A' < 0) ++j;
int diff = compressedString[i] - 'A';
string times = compressedString.substr(i + 1, j - i - 1);
q.push({diff, stoi(times)});
i = j;
}
}
char next() {
if (!hasNext()) return ' ';
vector<int>& t = q.front();
char ch = t[0] + 'A';
if (--t[1] == 0) q.pop();
return ch;
}
bool hasNext() {
return !q.empty();
}
private:
queue<vector<int> > q;
};Last updated
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