LeetCode 284. Peeking Iterator 题解

题目描述

给定迭代器Iterator支持next() 和 hasNext()操作，设计PeekingIterator迭代器，使其支持peek()操作。

解题思路

用一个外部变量保存next()要返回的下一个元素，调用peek()时直接返回该变量，调用next()方法时更新该外部变量。

代码实现

# Below is the interface for Iterator, which is already defined for you.
#
# class Iterator(object):
#     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(object):

    def __init__(self, iterator):
        """
        Initialize your data structure here.
        :type iterator: Iterator
        """

        self.iter = iterator
        self.tmp = self.iter.next() if self.iter.hasNext() else None

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """

        return self.tmp

    def next(self):
        """
        :rtype: int
        """

        res = self.tmp
        self.tmp = self.iter.next() if self.iter.hasNext() else None
        return res

    def hasNext(self):
        """
        :rtype: bool
        """

        return self.tmp is not None