leetcode/0284_peeking-iterator/python3/solution.py

66 lines
1.6 KiB
Python

# 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:
peeked = None
def __init__(self, iterator):
"""
Initialize your data structure here.
:type iterator: Iterator
"""
self.it = iterator
def peek(self):
"""
Returns the next element in the iteration without advancing the iterator.
:rtype: int
"""
if self.peeked:
return self.peeked
self.peeked = self.it.next()
return self.peeked
def next(self):
"""
:rtype: int
"""
if self.peeked:
tmp, self.peeked = self.peeked, None
return tmp
return self.it.next()
def hasNext(self):
"""
:rtype: bool
"""
return self.peeked is not None or self.it.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].