Design Compressed String Iterator
Similar Problems:
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 ' '
Github: code.dennyzhang.com
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/design-compressed-string-iterator ## Basic Ideas: ## ## Complexity: class StringIterator: def __init__(self, compressedString): """ :type compressedString: str """ self.string = compressedString self.index, self.count, self.ch = 0, 0, None self.length = len(self.string) def next(self): """ :rtype: str """ if self.hasNext() is False: return ' ' if self.count != 0: self.count -= 1 else: # fetch the next character self.ch = self.string[self.index] self.index += 1 # find the count v = '' while self.index < self.length: char = self.string[self.index] if char.isdigit() is False: break v = '%s%s' % (v, char) self.index += 1 self.count = int(v) - 1 return self.ch def hasNext(self): """ :rtype: bool """ if self.index == self.length and self.count == 0: return False else: return True # Your StringIterator object will be instantiated and called as such: # obj = StringIterator(compressedString) # param_1 = obj.next() # param_2 = obj.hasNext()
