284. Peeking Iterator

[zeikar]

Problem link

https://leetcode.com/problems/peeking-iterator/

Problem Summary

일반 이터레이터를 사용해서 peek 연산이 가능한 이터레이터를 만드는 문제.

Solution

임시 변수 하나를 두면 쉽게 풀린다.

peek 할 경우 peeked 변수에 값을 집어넣고, 이미 변수에 값이 있다면 그대로 리턴하면 된다.
next 할 때는 peeked 변수에 값이 있으면 그대로 리턴하면 되고 값이 없으면 next를 호출해서 가져오면 된다.
hasnext도 비슷하게 peeked 값이 있으면 true, 없으면 hasnext 호출하면 된다.

Source Code

# Below is the interface for Iterator, which is already defined for you.
#
class Iterator:
    def __init__(self, nums):
        """
        Initializes an iterator object to the beginning of a list.
        :type nums: List[int]
        """

    def hasNext(self):
        """
        Returns true if the iteration has more elements.
        :rtype: bool
        """

    def next(self):
        """
        Returns the next element in the iteration.
        :rtype: int
        """


class PeekingIterator:
    def __init__(self, iterator):
        """
        Initialize your data structure here.
        :type iterator: Iterator
        """
        self.iterator = iterator
        self.peeked = None

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """
        if self.peeked is None:
            self.peeked = self.iterator.next()
        return self.peeked

    def next(self):
        """
        :rtype: int
        """
        if self.peeked is not None:
            result = self.peeked
            self.peeked = None
            return result
        return self.iterator.next()

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.peeked is not None or self.iterator.hasNext()

# Your PeekingIterator object will be instantiated and called as such:
# iter = PeekingIterator(Iterator(nums))
# while iter.hasNext():
#     val = iter.peek()   # Get the next element but not advance the iterator.
#     iter.next()         # Should return the same value as [val].