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  • 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].
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  • 原文地址:https://www.cnblogs.com/xxxsans/p/13827792.html
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