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.

Follow up: How would you extend your design to be generic and work with all types, not just integer?

Credits:
Special thanks to @porker2008 for adding this problem and creating all test cases.

给了一个迭代器Iterator类接口有next(),hasNext()方法,让在此基础上设计实现一个PeekingIterator,有如下函数:

peek():返回下一个元素,但指针不移动到下一个
next(): 移动到下一个元素x并返回x
hasNext() :返回有下一个元素

解法:为了能peek()后下次next()还得到同样的数字,要用一个缓存保存下一个数字。当peek()时,返回缓存里的数就行,迭代器位置不会变。当next()的时候除了要返回数字和指针移动一位,还要将缓存更新为下一个数字,如果没有下一个就将缓存更新为null。

Java:

class PeekingIterator implements Iterator<Integer> {
private Iterator<Integer> iter;
private Integer nextElement;
private boolean peekUsed; public PeekingIterator(Iterator<Integer> iterator) {
// initialize any member here.
iter = iterator;
} // Returns the next element in the iteration without advancing the iterator.
public Integer peek() {
if(!peekUsed) {
nextElement = iter.next();
peekUsed = true;
}
return nextElement;
} // hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
@Override
public Integer next() {
if(peekUsed) {
peekUsed = false;
return nextElement;
}
return iter.next();
} @Override
public boolean hasNext() {
if(peekUsed) {
return true;
}
return iter.hasNext();
}
}

Java:  用Integer来做global variable, 不用使用boolean

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
private Iterator<Integer> iter;
private Integer next; public PeekingIterator(Iterator<Integer> iterator) {
// initialize any member here.
iter = iterator;
if (iter.hasNext()) {
next = iter.next();
}
} // Returns the next element in the iteration without advancing the iterator.
public Integer peek() {
return next;
} // hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
@Override
public Integer next() {
Integer res = next;
next = iter.hasNext() ? iter.next() : null;
return res;
} @Override
public boolean hasNext() {
return next != null;
}
}

Python:

class PeekingIterator(object):
def __init__(self, iterator):
"""
Initialize your data structure here.
:type iterator: Iterator
"""
self.iterator = iterator
self.val_ = None
self.has_next_ = iterator.hasNext()
self.has_peeked_ = False def peek(self):
"""
Returns the next element in the iteration without advancing the iterator.
:rtype: int
"""
if not self.has_peeked_:
self.has_peeked_ = True
self.val_ = self.iterator.next()
return self.val_; def next(self):
"""
:rtype: int
"""
self.val_ = self.peek()
self.has_peeked_ = False
self.has_next_ = self.iterator.hasNext()
return self.val_; def hasNext(self):
"""
:rtype: bool
"""
return self.has_next_   

C++:

// Below is the interface for Iterator, which is already defined for you.
// **DO NOT** modify the interface for Iterator.
class Iterator {
struct Data;
Data* data;
public:
Iterator(const vector<int>& nums);
Iterator(const Iterator& iter);
virtual ~Iterator();
// Returns the next element in the iteration.
int next();
// Returns true if the iteration has more elements.
bool hasNext() const;
}; class PeekingIterator : public Iterator {
public:
PeekingIterator(const vector<int>& nums) : Iterator(nums) {
// Initialize any member here.
// **DO NOT** save a copy of nums and manipulate it directly.
// You should only use the Iterator interface methods.
_flag = false;
} // Returns the next element in the iteration without advancing the iterator.
int peek() {
if (!_flag) {
_value = Iterator::next();
_flag = true;
}
return _value;
} // hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
int next() {
if (!_flag) return Iterator::next();
_flag = false;
return _value;
} bool hasNext() const {
if (_flag) return true;
if (Iterator::hasNext()) return true;
return false;
}
private:
int _value;
bool _flag;
};

  

All LeetCode Questions List 题目汇总

[LeetCode] 284. Peeking Iterator 瞥一眼迭代器的更多相关文章

  1. 【LeetCode】284. Peeking Iterator 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/peeking-i ...

  2. 【LeetCode】284. Peeking Iterator

    题目: Given an Iterator class interface with methods: next() and hasNext(), design and implement a Pee ...

  3. [LeetCode] 281. Zigzag Iterator 之字形迭代器

    Given two 1d vectors, implement an iterator to return their elements alternately. Example: Input: v1 ...

  4. 284. Peeking Iterator

    题目: Given an Iterator class interface with methods: next() and hasNext(), design and implement a Pee ...

  5. 设计模式 - 迭代模式(iterator pattern) Java 迭代器(Iterator) 详细解释

    迭代模式(iterator pattern) Java 迭代器(Iterator) 详细解释 本文地址: http://blog.csdn.net/caroline_wendy 參考迭代器模式(ite ...

  6. Java容器类源码分析之Iterator与ListIterator迭代器(基于JDK8)

    一.基本概念 迭代器是一个对象,也是一种设计模式,Java有两个用来实实现迭代器的接口,分别是Iterator接口和继承自Iterator的ListIterator接口.实现迭代器接口的类的对象有遍历 ...

  7. Collection接口都是通过Iterator()(即迭代器)来对Set和List遍历

    以下介绍接口: List接口:(介绍其下的两个实现类:ArrayList和LinkedList) ArrayList和数组非常类似,其底层①也用数组组织数据,ArrayList是动态可变数组. ① 底 ...

  8. [LeetCode] Peeking Iterator 顶端迭代器

    Given an Iterator class interface with methods: next() and hasNext(), design and implement a Peeking ...

  9. LeetCode OJ:Peeking Iterator(peeking 迭代器)

    Given an Iterator class interface with methods: next() and hasNext(), design and implement a Peeking ...

随机推荐

  1. 《exception》第九次团队作业:Beta冲刺与验收准备(第三天)

    一.项目基本介绍 项目 内容 这个作业属于哪个课程 任课教师博客主页链接 这个作业的要求在哪里 作业链接地址 团队名称 Exception 作业学习目标 1.掌握软件黑盒测试技术:2.学会编制软件项目 ...

  2. Java 15周作业

    题目1:编写一个应用程序,输入用户名和密码,访问test数据库中t_login表(字段包括id.username.password),验证登录是否成功. 题目2:在上一题基础上,当登录成功后,将t_u ...

  3. django-改写manage类-objects

    user/models.py中 class AddressManage(models.Manager): '''地址模型管理类''' def get_default_addr(self, user): ...

  4. hdu3625

    hdu3625 题意: 酒店发生一起谋杀案.作为镇上最好的侦探,您应该立即检查酒店的所有N个房间.但是,房间的所有门都是锁着的,钥匙刚锁在房间里,真是个陷阱!您知道每个房间里只有一把钥匙,并且所有可能 ...

  5. 关于原生js的节点兼容性

    关于节点的兼容性: 1:获取元素的子节点 a: childNodes:获取元素的子节点,空文本,非空文本,注释,获取的比较全面, 如果只是想获取元素的子节点,请用(children) b:     c ...

  6. linux学习8 运维基本功-Linux获取命令使用帮助详解

    一.Linux基础知识 1.人机交互界面: a.GUI b.CLI:[login@hostname workdir]# COMMAND 2.命令知识 通用格式:# COMMAND  OPTIONS A ...

  7. Windbg命令脚本

    命令脚本,就是将完成某个特定任务的相关命令组合在一起,保存在脚本文件里,加载到Windbg里执行,达到我们的目的.你可以理解为脚本就是一种语言,就像c或者汇编,但是他不需要编译器将其编译为可执行文件, ...

  8. Lightning Web Components 组合(五)

    使用组合我们可以用来设计复杂的组件. 组合一些比较小的组件,可以增加组件的重新性以及可维护性. 通过以下一个简单的demo,将会展示关于owner 以及container 的概念,在实际的项目中 ex ...

  9. dinoql 使用nodejs 运行的几个问题

    dinoql 是一个很不错的javascript objects 查询处理方案,基于graphql,当前版本有点问题 node 环境运行 ReferenceError: window is not d ...

  10. SP19997 MOON2 - Moon Safari (Hard) 【数论,多项式】

    题目描述:求 \[ \sum_{i=1}^ni^kr^i \] 对某个质数取模.\(T\)组数据. 数据范围:\(n,r\le 10^{18},\sum k\le 2.56\times 10^6\) ...