360 lines
10 KiB
Markdown
360 lines
10 KiB
Markdown
<p align="center">
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<a href="https://mp.weixin.qq.com/s/RsdcQ9umo09R6cfnwXZlrQ"><img src="https://img.shields.io/badge/PDF下载-代码随想录-blueviolet" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw"><img src="https://img.shields.io/badge/刷题-微信群-green" alt=""></a>
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<a href="https://space.bilibili.com/525438321"><img src="https://img.shields.io/badge/B站-代码随想录-orange" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ"><img src="https://img.shields.io/badge/知识星球-代码随想录-blue" alt=""></a>
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</p>
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<p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!</strong></p>
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> 工作上一定没人这么搞,但是考察对栈、队列理解程度的好题
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# 232.用栈实现队列
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[力扣题目链接](https://leetcode-cn.com/problems/implement-queue-using-stacks/)
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使用栈实现队列的下列操作:
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push(x) -- 将一个元素放入队列的尾部。
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pop() -- 从队列首部移除元素。
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peek() -- 返回队列首部的元素。
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empty() -- 返回队列是否为空。
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示例:
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```
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MyQueue queue = new MyQueue();
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queue.push(1);
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queue.push(2);
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queue.peek(); // 返回 1
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queue.pop(); // 返回 1
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queue.empty(); // 返回 false
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```
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说明:
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* 你只能使用标准的栈操作 -- 也就是只有 push to top, peek/pop from top, size, 和 is empty 操作是合法的。
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* 你所使用的语言也许不支持栈。你可以使用 list 或者 deque(双端队列)来模拟一个栈,只要是标准的栈操作即可。
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* 假设所有操作都是有效的 (例如,一个空的队列不会调用 pop 或者 peek 操作)。
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## 思路
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这是一道模拟题,不涉及到具体算法,考察的就是对栈和队列的掌握程度。
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使用栈来模式队列的行为,如果仅仅用一个栈,是一定不行的,所以需要两个栈**一个输入栈,一个输出栈**,这里要注意输入栈和输出栈的关系。
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下面动画模拟以下队列的执行过程如下:
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执行语句:
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queue.push(1);
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queue.push(2);
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queue.pop(); **注意此时的输出栈的操作**
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queue.push(3);
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queue.push(4);
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queue.pop();
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queue.pop();**注意此时的输出栈的操作**
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queue.pop();
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queue.empty();
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在push数据的时候,只要数据放进输入栈就好,**但在pop的时候,操作就复杂一些,输出栈如果为空,就把进栈数据全部导入进来(注意是全部导入)**,再从出栈弹出数据,如果输出栈不为空,则直接从出栈弹出数据就可以了。
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最后如何判断队列为空呢?**如果进栈和出栈都为空的话,说明模拟的队列为空了。**
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在代码实现的时候,会发现pop() 和 peek()两个函数功能类似,代码实现上也是类似的,可以思考一下如何把代码抽象一下。
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C++代码如下:
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```CPP
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class MyQueue {
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public:
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stack<int> stIn;
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stack<int> stOut;
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/** Initialize your data structure here. */
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MyQueue() {
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}
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/** Push element x to the back of queue. */
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void push(int x) {
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stIn.push(x);
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}
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/** Removes the element from in front of queue and returns that element. */
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int pop() {
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// 只有当stOut为空的时候,再从stIn里导入数据(导入stIn全部数据)
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if (stOut.empty()) {
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// 从stIn导入数据直到stIn为空
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while(!stIn.empty()) {
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stOut.push(stIn.top());
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stIn.pop();
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}
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}
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int result = stOut.top();
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stOut.pop();
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return result;
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}
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/** Get the front element. */
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int peek() {
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int res = this->pop(); // 直接使用已有的pop函数
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stOut.push(res); // 因为pop函数弹出了元素res,所以再添加回去
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return res;
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}
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/** Returns whether the queue is empty. */
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bool empty() {
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return stIn.empty() && stOut.empty();
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}
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};
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```
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## 拓展
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可以看出peek()的实现,直接复用了pop()。
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再多说一些代码开发上的习惯问题,在工业级别代码开发中,最忌讳的就是 实现一个类似的函数,直接把代码粘过来改一改就完事了。
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这样的项目代码会越来越乱,**一定要懂得复用,功能相近的函数要抽象出来,不要大量的复制粘贴,很容易出问题!(踩过坑的人自然懂)**
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工作中如果发现某一个功能自己要经常用,同事们可能也会用到,自己就花点时间把这个功能抽象成一个好用的函数或者工具类,不仅自己方便,也方面了同事们。
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同事们就会逐渐认可你的工作态度和工作能力,自己的口碑都是这么一点一点积累起来的!在同事圈里口碑起来了之后,你就发现自己走上了一个正循环,以后的升职加薪才少不了你!哈哈哈
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## 其他语言版本
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Java:
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```java
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class MyQueue {
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Stack<Integer> stack1;
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Stack<Integer> stack2;
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/** Initialize your data structure here. */
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public MyQueue() {
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stack1 = new Stack<>(); // 负责进栈
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stack2 = new Stack<>(); // 负责出栈
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}
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/** Push element x to the back of queue. */
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public void push(int x) {
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stack1.push(x);
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}
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/** Removes the element from in front of queue and returns that element. */
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public int pop() {
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dumpStack1();
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return stack2.pop();
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}
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/** Get the front element. */
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public int peek() {
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dumpStack1();
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return stack2.peek();
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}
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/** Returns whether the queue is empty. */
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public boolean empty() {
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return stack1.isEmpty() && stack2.isEmpty();
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}
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// 如果stack2为空,那么将stack1中的元素全部放到stack2中
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private void dumpStack1(){
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if (stack2.isEmpty()){
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while (!stack1.isEmpty()){
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stack2.push(stack1.pop());
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}
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}
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}
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}
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/**
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* Your MyQueue object will be instantiated and called as such:
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* MyQueue obj = new MyQueue();
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* obj.push(x);
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* int param_2 = obj.pop();
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* int param_3 = obj.peek();
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* boolean param_4 = obj.empty();
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*/
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```
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Python:
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```python
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class MyQueue:
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def __init__(self):
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"""
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in主要负责push,out主要负责pop
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"""
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self.stack_in = []
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self.stack_out = []
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def push(self, x: int) -> None:
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"""
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有新元素进来,就往in里面push
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"""
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self.stack_in.append(x)
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def pop(self) -> int:
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"""
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Removes the element from in front of queue and returns that element.
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"""
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if self.empty():
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return None
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if self.stack_out:
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return self.stack_out.pop()
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else:
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for i in range(len(self.stack_in)):
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self.stack_out.append(self.stack_in.pop())
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return self.stack_out.pop()
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def peek(self) -> int:
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"""
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Get the front element.
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"""
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ans = self.pop()
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self.stack_out.append(ans)
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return ans
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def empty(self) -> bool:
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"""
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只要in或者out有元素,说明队列不为空
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"""
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return not (self.stack_in or self.stack_out)
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```
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Go:
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```Go
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type MyQueue struct {
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stack []int
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back []int
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}
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/** Initialize your data structure here. */
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func Constructor() MyQueue {
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return MyQueue{
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stack: make([]int, 0),
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back: make([]int, 0),
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}
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}
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/** Push element x to the back of queue. */
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func (this *MyQueue) Push(x int) {
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for len(this.back) != 0 {
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val := this.back[len(this.back)-1]
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this.back = this.back[:len(this.back)-1]
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this.stack = append(this.stack, val)
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}
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this.stack = append(this.stack, x)
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}
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/** Removes the element from in front of queue and returns that element. */
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func (this *MyQueue) Pop() int {
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for len(this.stack) != 0 {
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val := this.stack[len(this.stack)-1]
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this.stack = this.stack[:len(this.stack)-1]
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this.back = append(this.back, val)
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}
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if len(this.back) == 0 {
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return 0
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}
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val := this.back[len(this.back)-1]
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this.back = this.back[:len(this.back)-1]
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return val
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}
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/** Get the front element. */
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func (this *MyQueue) Peek() int {
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for len(this.stack) != 0 {
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val := this.stack[len(this.stack)-1]
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this.stack = this.stack[:len(this.stack)-1]
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this.back = append(this.back, val)
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}
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if len(this.back) == 0 {
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return 0
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}
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val := this.back[len(this.back)-1]
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return val
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}
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/** Returns whether the queue is empty. */
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func (this *MyQueue) Empty() bool {
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return len(this.stack) == 0 && len(this.back) == 0
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}
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```
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javaScript:
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```js
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// 使用两个数组的栈方法(push, pop) 实现队列
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/**
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* Initialize your data structure here.
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*/
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var MyQueue = function() {
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this.stack1 = [];
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this.stack2 = [];
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};
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/**
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* Push element x to the back of queue.
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* @param {number} x
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* @return {void}
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*/
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MyQueue.prototype.push = function(x) {
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this.stack1.push(x);
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};
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/**
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* Removes the element from in front of queue and returns that element.
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* @return {number}
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*/
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MyQueue.prototype.pop = function() {
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const size = this.stack2.length;
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if(size) {
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return this.stack2.pop();
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}
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while(this.stack1.length) {
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this.stack2.push(this.stack1.pop());
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}
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return this.stack2.pop();
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};
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/**
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* Get the front element.
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* @return {number}
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*/
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MyQueue.prototype.peek = function() {
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const x = this.pop();
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this.stack2.push(x);
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return x;
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};
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/**
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* Returns whether the queue is empty.
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* @return {boolean}
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*/
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MyQueue.prototype.empty = function() {
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return !this.stack1.length && !this.stack2.length
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};
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```
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-----------------------
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* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频:[代码随想录](https://space.bilibili.com/525438321)
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* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>
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