题目描述：
Given an Iterator class interface with methods: next() and hasNext(), design and implement a PeekingIterator that support the peek() operation -- it essentially peek() at the element that will be returned by the next call to next().
Here is an example. Assume that the iterator is initialized to the beginning of the list: [1, 2, 3].
Call next() gets you 1, the first element in the list.
Now you call peek() and it returns 2, the next element. Calling next() after that still return 2.
You call next() the final time and it returns 3, the last element. Calling hasNext() after that should return false.
Hint:
Think of "looking ahead". You want to cache the next element.
Is one variable sufficient? Why or why not?
Test your design with call order of peek() before next() vs next() before peek().
For a clean implementation, check out Google's guava library source code.
Follow up: How would you extend your design to be generic and work with all types, not just integer?
题目大意：
给定一个迭代器类接口，包含方法: next()和hasNext()，设计并实现一个PeekingIterator支持peek()操作 -- 其本质就是把原本应由next()方法返回的元素peek()出来。
这里有一个例子。假设迭代器初始化为list：[1, 2, 3]。
调用next()得到1，列表的第一个元素。
现在调用peek()然后返回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。此后调用hasNext()应该返回false。
提示：
考虑"预先获取"。你可能需要缓存下一个元素。
一个变量够用吗？原因是什么？
通过在next()之前调用peek()，以及在peek()之前调用next()测试你的设计。
关于本问题的一个清晰的实现，可以参考Google的guava库源代码。
进一步思考：怎样拓展你的设计？使之变得通用化，适应所有的类型，而不只是整数？
解题思路：
本题目考察设计模式中的装饰器模式（Decorator Pattern）
参阅维基百科Decorator Pattern词条：https://en.wikipedia.org/wiki/Decorator_pattern
In object-oriented programming, the decorator pattern (also known as Wrapper, an alternative naming shared with the Adapter pattern) is a design pattern that allows behavior to be added to an individual object, either statically or dynamically, without affecting the behavior of other objects from the same class. The decorator pattern is often useful for adhering to the Single Responsibility Principle, as it allows functionality to be divided between classes with unique areas of concern.
引入两个额外的变量nextElement和peekFlag
nextElement标识peek操作预先获取的下一个元素，peekFlag记录当前是否已经执行过peek操作
若已知所有元素非空，不使用peekFlag变量也可，参考：https://leetcode.com/discuss/59327/my-java-solution
关于进一步思考，使用Java的泛型可以适用于所有的类型。
具体实现详见代码。
Python代码：
# 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.peekFlag = False
self.nextElement = None
def peek(self):
"""
Returns the next element in the iteration without advancing the iterator.
:rtype: int
"""
if not self.peekFlag:
self.nextElement = self.iter.next()
self.peekFlag = True
return self.nextElement
def next(self):
"""
:rtype: int
"""
if not self.peekFlag:
return self.iter.next()
nextElement = self.nextElement
self.peekFlag = False
self.nextElement = None
return nextElement
def hasNext(self):
"""
:rtype: bool
"""
return self.peekFlag or self.iter.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].
Java代码：
// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
private Iterator<Integer> iterator;
private boolean peekFlag = false;
private Integer nextElement = null;
public PeekingIterator(Iterator<Integer> iterator) {
// initialize any member here.
this.iterator = iterator;
}
// Returns the next element in the iteration without advancing the iterator.
public Integer peek() {
if (!peekFlag) {
nextElement = iterator.next();
peekFlag = true;
}
return nextElement;
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
@Override
public Integer next() {
if (!peekFlag) {
return iterator.next();
}
peekFlag = false;
Integer result = nextElement;
nextElement = null;
return result;
}
@Override
public boolean hasNext() {
return peekFlag || iterator.hasNext();
}
}
本文链接：http://bookshadow.com/weblog/2015/09/21/leetcode-peeking-iterator/
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