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窥视迭代器
请你在设计一个迭代器，在集成现有迭代器拥有的 hasNext 和 next 操作的基础上，还额外支持 peek 操作。 实现 PeekingIterator 类： PeekingIterator(Iterator nums) 使用指定整数迭代器 nums 初始化迭代器。 int next() 返回数…
3
题型
4
代码语言
3
相关题
当前训练重点
中等 · 数组·结合·design
答案摘要
class PeekingIterator: def __init__(self, iterator):
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题目描述
请你在设计一个迭代器，在集成现有迭代器拥有的 hasNext 和 next 操作的基础上，还额外支持 peek 操作。
实现 PeekingIterator 类：
PeekingIterator(Iterator<int> nums)使用指定整数迭代器nums初始化迭代器。int next()返回数组中的下一个元素，并将指针移动到下个元素处。bool hasNext()如果数组中存在下一个元素，返回true；否则，返回false。int peek()返回数组中的下一个元素，但 不 移动指针。
注意：每种语言可能有不同的构造函数和迭代器 Iterator，但均支持 int next() 和 boolean hasNext() 函数。
示例 1：
输入： ["PeekingIterator", "next", "peek", "next", "next", "hasNext"] [[[1, 2, 3]], [], [], [], [], []] 输出： [null, 1, 2, 2, 3, false] 解释： PeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [1,2,3] peekingIterator.next(); // 返回 1 ，指针移动到下一个元素 [1,2,3] peekingIterator.peek(); // 返回 2 ，指针未发生移动 [1,2,3] peekingIterator.next(); // 返回 2 ，指针移动到下一个元素 [1,2,3] peekingIterator.next(); // 返回 3 ，指针移动到下一个元素 [1,2,3] peekingIterator.hasNext(); // 返回 False
提示：
1 <= nums.length <= 10001 <= nums[i] <= 1000- 对
next和peek的调用均有效 next、hasNext和peek最多调用1000次
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？
解题思路
方法一
# 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.has_peeked = False
self.peeked_element = None
def peek(self):
"""
Returns the next element in the iteration without advancing the iterator.
:rtype: int
"""
if not self.has_peeked:
self.peeked_element = self.iterator.next()
self.has_peeked = True
return self.peeked_element
def next(self):
"""
:rtype: int
"""
if not self.has_peeked:
return self.iterator.next()
result = self.peeked_element
self.has_peeked = False
self.peeked_element = None
return result
def hasNext(self):
"""
:rtype: bool
"""
return self.has_peeked or 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].
复杂度分析
| 指标 | 值 |
|---|---|
| 时间 | Depends on the final approach |
| 空间 | Depends on the final approach |
面试官常问的追问
外企场景- question_mark
Look for understanding of iterator design patterns and the need for efficient caching.
- question_mark
Test the candidate's ability to handle edge cases like empty or fully traversed iterators.
- question_mark
Gauge how well the candidate manages state within the iterator, ensuring peek() doesn't disrupt the iterator's flow.
常见陷阱
外企场景- error
Failing to properly cache the next element, leading to incorrect results from peek().
- error
Not handling edge cases like calling next or peek when there are no remaining elements.
- error
Overcomplicating the solution by storing unnecessary state, increasing space complexity.
进阶变体
外企场景- arrow_right_alt
Designing an iterator with more advanced peek features, such as peeking multiple steps ahead.
- arrow_right_alt
Adding support for resetting the iterator back to the start after completing the iteration.
- arrow_right_alt
Expanding the iterator's functionality to support different data types, like strings or custom objects.