mirror of
https://github.com/chefyuan/algorithm-base.git
synced 2024-11-24 21:08:53 +00:00
82 lines
2.5 KiB
Markdown
82 lines
2.5 KiB
Markdown
|
今天给大家带来一个有意思的题目,思路很easy,但是刚刷题的小伙伴,示例理解起来可能会有点费劲,花里胡哨一大堆是啥意思啊。在之前的文章《不知道这篇文章合不合你的胃口》中写了栈是先进后出,队列是先进先出。本题让我们用两个先进后出的栈,完成一个先进先出的队列。我们应该怎么实现呢?
|
|||
|
|
|||
|
废话不多说,大家看图
|
|||
|
|
|||
|
![栈实现队列](E:\Typora笔记\CSDN\leetcode通关笔记\博客动图\栈实现队列.gif)
|
|||
|
|
|||
|
这就是具体思路,然后我们来看一下题目示例及官方提供的函数都是什么意思。
|
|||
|
|
|||
|
```java
|
|||
|
class CQueue {
|
|||
|
//创建队列
|
|||
|
public CQueue() {
|
|||
|
|
|||
|
}
|
|||
|
//入队
|
|||
|
public void appendTail(int value) {
|
|||
|
|
|||
|
}
|
|||
|
//出队
|
|||
|
public int deleteHead() {
|
|||
|
|
|||
|
}
|
|||
|
}
|
|||
|
```
|
|||
|
|
|||
|
示例 1:
|
|||
|
|
|||
|
```java
|
|||
|
输入:
|
|||
|
["CQueue","appendTail","deleteHead","deleteHead"]
|
|||
|
[[],[3],[],[]]
|
|||
|
输出:[null,null,3,-1]
|
|||
|
```
|
|||
|
|
|||
|
|
|||
|
示例 2:
|
|||
|
|
|||
|
```java
|
|||
|
输入:
|
|||
|
["CQueue","deleteHead","appendTail","appendTail","deleteHead","deleteHead"]
|
|||
|
[[],[],[5],[2],[],[]]
|
|||
|
输出:[null,-1,null,null,5,2]
|
|||
|
```
|
|||
|
|
|||
|
其实也很容易理解,输入有两行第一行,为执行的函数,Cqueue代表创建队列(代表我们初始化两个栈),appendTail代表入队操作(代表stackA入栈),deleteHead代表出队操作(代表我们stackB出栈)
|
|||
|
|
|||
|
第二行输入代表值,分别给每个函数传入的参数,我们发现只有appendTail函数下有对应值,因为只有该函数传入参数。
|
|||
|
|
|||
|
大家可以点击该链接[剑指 Offer 09. 用两个栈实现队列](https://leetcode-cn.com/problems/yong-liang-ge-zhan-shi-xian-dui-lie-lcof/)去实现一下,下面我们看代码。
|
|||
|
|
|||
|
```java
|
|||
|
class CQueue {
|
|||
|
//初始化两个栈
|
|||
|
Stack<Integer> stack1,stack2;
|
|||
|
public CQueue() {
|
|||
|
stack1 = new Stack<>();
|
|||
|
stack2 = new Stack<>();
|
|||
|
|
|||
|
}
|
|||
|
//入队,我们往第一个栈压入值
|
|||
|
public void appendTail (int value) {
|
|||
|
stack1.push(value);
|
|||
|
}
|
|||
|
//出队
|
|||
|
public int deleteHead() {
|
|||
|
//大家可以自己思考一下为什么if条件为stack2.isEmpty(),细节所在
|
|||
|
if (stack2.isEmpty()) {
|
|||
|
//如果此时A栈没有值,则直接-1,我们可以看示例
|
|||
|
if (stack1.isEmpty()) {
|
|||
|
return -1;
|
|||
|
}
|
|||
|
//将A栈的值,压入B栈中
|
|||
|
while (!stack1.isEmpty()) {
|
|||
|
stack2.push(stack1.pop());
|
|||
|
}
|
|||
|
}
|
|||
|
return stack2.pop();
|
|||
|
}
|
|||
|
}
|
|||
|
```
|
|||
|
|