284. Peeking Iterator
https://leetcode.com/problems/peeking-iterator/
Level: Medium
Not much to explain. This problem is very easy.
2 ways to do this.
- Iterate and save all of iterator's value, and implement the functions
- Method 2 iterates through input iterator while itself is being iterated
- This type of solution can avoid saving a copy of values of the original iterator (not the current version, I will implement it)
# 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.ptr = 0
self.array = []
def peek(self):
"""
Returns the next element in the iteration without advancing the iterator.
:rtype: int
"""
if self.ptr >= len(self.array):
self.array.append(self.iterator.next())
return self.array[self.ptr]
def next(self):
"""
:rtype: int
"""
if self.ptr >= len(self.array):
self.array.append(self.iterator.next())
self.ptr += 1
return self.array[self.ptr - 1]
def hasNext(self):
"""
:rtype: bool
"""
return self.iterator.hasNext() or self.ptr < len(self.array)
# 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].