Leetcode: Design Compressed String Iterator

Posted on February 11, 2018September 23, 2019 by braindenny

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.

## 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()

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