LeetCode232 Implement Queue using Stacks Java 题解
题目:
Implement the following operations of a queue using stacks.
- push(x) -- Push element x to the back of queue.
- pop() -- Removes the element from in front of queue.
- peek() -- Get the front element.
- empty() -- Return whether the queue is empty.
Notes:
- You must use only standard operations of a stack -- which means only
push,
to toppeek/pop from top,size,
andis emptyoperations are valid. - Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
- You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
解题:
用栈实现队列,这个比用队列实现栈要麻烦一些,这里用到两个栈,两种思路。第一种就是在进栈的时候,把栈逆序放在另外一个栈,出的时候直接出另外一个栈就能够了。另外一种思路。进栈的时候不做不论什么处理。出栈的时候把栈逆序放在另外一个栈。出另外一个栈。以下就是两种的代码实现。
在进栈的时候进行处理:
class MyQueue {
// Push element x to the back of queue.
Stack<Integer> stack=new Stack<>();
Stack<Integer> stack2=new Stack<>();
public void push(int x) {
while(!stack.isEmpty())
{
stack2.push(stack.pop());
}
stack2.push(x);
while(!stack2.isEmpty())
{
stack.push(stack2.pop());
}
}
// Removes the element from in front of queue.
public void pop() {
stack.pop();
}
// Get the front element.
public int peek() {
return stack.peek();
}
// Return whether the queue is empty.
public boolean empty() {
return stack.isEmpty();
}
}
在出栈的时候进行处理:
class MyQueue2 {
// Push element x to the back of queue.
Stack<Integer> stack=new Stack<>();
Stack<Integer> stack2=new Stack<>();
public void push(int x) {
while(!stack2.isEmpty())
stack.push(stack2.pop());
stack.push(x);
}
// Removes the element from in front of queue.
public void pop() {
while(!stack.isEmpty())
stack2.push(stack.pop());
stack2.pop();
}
// Get the front element.
public int peek() {
while(!stack.isEmpty())
stack2.push(stack.pop());
return stack2.peek();
}
// Return whether the queue is empty.
public boolean empty() {
while(!stack2.isEmpty())
stack.push(stack2.pop());
return stack.isEmpty();
}
}
LeetCode232 Implement Queue using Stacks Java 题解的更多相关文章
- LeetCode232:Implement Queue using Stacks
Implement the following operations of a queue using stacks. push(x) – Push element x to the back of ...
- Lintcode: Implement Queue by Stacks 解题报告
Implement Queue by Stacks 原题链接 : http://lintcode.com/zh-cn/problem/implement-queue-by-stacks/# As th ...
- LeetCode 232. 用栈实现队列(Implement Queue using Stacks) 4
232. 用栈实现队列 232. Implement Queue using Stacks 题目描述 使用栈实现队列的下列操作: push(x) -- 将一个元素放入队列的尾部. pop() -- 从 ...
- leetcode:Implement Stack using Queues 与 Implement Queue using Stacks
一.Implement Stack using Queues Implement the following operations of a stack using queues. push(x) - ...
- 【LeetCode】232 & 225 - Implement Queue using Stacks & Implement Stack using Queues
232 - Implement Queue using Stacks Implement the following operations of a queue using stacks. push( ...
- 232. Implement Queue using Stacks,225. Implement Stack using Queues
232. Implement Queue using Stacks Total Accepted: 27024 Total Submissions: 79793 Difficulty: Easy Im ...
- leetcode 155. Min Stack 、232. Implement Queue using Stacks 、225. Implement Stack using Queues
155. Min Stack class MinStack { public: /** initialize your data structure here. */ MinStack() { } v ...
- Leetcode 232 Implement Queue using Stacks 和 231 Power of Two
1. 232 Implement Queue using Stacks 1.1 问题描写叙述 使用栈模拟实现队列.模拟实现例如以下操作: push(x). 将元素x放入队尾. pop(). 移除队首元 ...
- 【LeetCode】232. Implement Queue using Stacks 解题报告(Python & Java)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 Python解法 Java解法 日期 [LeetCo ...
随机推荐
- Android兼容性测试GTS-环境搭建、测试执行、结果分析
GTS的全称是Google Mobile Services Test Suite,所谓的Google Mobile Services即谷歌移动服务,是谷歌开发并推动Android的动力,也是Andro ...
- django的rest framework框架——安装及基本使用
一.django的FBV 和 CBV 1.FBV(基于函数的视图): urlpatterns = [ url(r'^users/', views.users), ] def users(request ...
- HDU 5016 Mart Master II
Mart Master II Time Limit: 6000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ...
- 2017ACM/ICPC广西邀请赛-重现赛(感谢广西大学)
上一场CF打到心态爆炸,这几天也没啥想干的 A Math Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/3 ...
- 【JavaScript 14—学习总结】:从小事做起
导读:花了将近两个月,JavaScript的学习视频算是做完了.里面的例子,都敲过一遍,但有少数的几个就是实现不了,比如:百度分享侧栏随着滚动条移动:菜单切换只对第一个起作用等,也就先放着了.现在,就 ...
- 九度oj 题目1342:寻找最长合法括号序列II
题目描述: 假如给你一个由’(‘和’)’组成的一个随机的括号序列,当然,这个括号序列肯定不能保证是左右括号匹配的,所以给你的任务便是去掉其中的一些括号,使得剩下的括号序列能够左右括号匹配且长度最长,即 ...
- ios弹性头部
很久没写博客了,金天有点时间来写下,一直觉得弹性头部很炫,看起来高大上,写起来蛮简单的 层次分析 一共有3层,最底部是图像层,中间是scrollView或者它的子类,最上层是scrollView上面添 ...
- 2016 ACM-ICPC China Finals #F Mr. Panda and Fantastic Beasts
题目链接$\newcommand{\LCP}{\mathrm{LCP}}\newcommand{\suf}{\mathrm{suf}}$ 题意 给定 $n$ 个字符串 $s_1, s_2, \dots ...
- [BZOJ1576] [Usaco2009 Jan]安全路经Travel(堆优化dijk + (并查集 || 树剖))
传送门 蒟蒻我原本还想着跑两边spfa,发现不行,就gg了. 首先这道题卡spfa,所以需要用堆优化的dijkstra求出最短路径 因为题目中说了,保证最短路径有且只有一条,所以可以通过dfs求出最短 ...
- Unslider--入门篇
Unslider--入门篇 背景:因工作需求,需要完成一个图片轮播效果,因博主不是专业的前端开发人员,so google之,经过挑选最终选择使用Unslider插件完成工作. 一.Unslider插件 ...