284. 顶端迭代器

给定一个迭代器类的接口，接口包含两个方法： next() 和 hasNext()。设计并实现一个支持 peek() 操作的顶端迭代器 -- 其本质就是把原本应由 next() 方法返回的元素 peek() 出来。

示例:

假设迭代器被初始化为列表 [1,2,3]。

调用 next() 返回 1，得到列表中的第一个元素。
现在调用 peek() 返回 2，下一个元素。在此之后调用 next() 仍然返回 2。
最后一次调用 next() 返回 3，末尾元素。在此之后调用 hasNext() 应该返回 false。
进阶：你将如何拓展你的设计？使之变得通用化，从而适应所有的类型，而不只是整数型？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/peeking-iterator
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

# 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.peek_num=None
        

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

    def next(self):
        """
        :rtype: int
        """
        if self.peek_num==None:
            return self.iterator.next()
        temp=self.peek_num
        self.peek_num=None
        return temp
        

    def hasNext(self):
        """
        :rtype: bool
        """
        if self.peek_num:
            return True
        return 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].