如何使用兩個(gè)堆棧實(shí)現(xiàn)隊(duì)列

Stack and Queue at a glance...

堆疊和排隊(duì)一目了然...

Stack:

堆棧：

The stack is an ordered list where insertion and deletion are done from the same end, top. The last element that entered first is the first one to be deleted (the basic principle behind the LIFO). That means it is a data structure which is implemented as LIFO.

堆棧是一個(gè)有序列表，其中插入和刪除從同一末端開(kāi)始。 首先輸入的最后一個(gè)元素是要?jiǎng)h除的第一個(gè)元素(LIFO的基本原理)。 這意味著它是一個(gè)實(shí)現(xiàn)為L(zhǎng)IFO的數(shù)據(jù)結(jié)構(gòu)。

The main stack operations are (basic ADT operations)...

主堆棧操作為(基本ADT操作)...

push (int data): Insertion at top

push(int數(shù)據(jù))：在頂部插入

int pop(): Deletion from top

int pop()：從頂部刪除

Queue:

隊(duì)列：

The queue is an ordered list in which insertions are done at one end (rear) and deletions are done from another end (front). The first element that got inserted is the first one to be deleted (basic principle of FIFO). That means it is a data structure which is implemented as FIFO.

隊(duì)列是一個(gè)有序列表，其中插入是在一端(后部)完成，而刪除是從另一端(前部)完成。 插入的第一個(gè)元素是要?jiǎng)h除的第一個(gè)元素(FIFO的基本原理)。 這意味著它是一個(gè)實(shí)現(xiàn)為FIFO的數(shù)據(jù)結(jié)構(gòu)。

The main Queue operations are (basic ADT operations)...

主要的Queue操作是(基本ADT操作)...

EnQueue (int data): Insertion at rear end

EnQueue(int數(shù)據(jù))：在后端插入

int DeQueue(): Deletion from front end

int DeQueue()：從前端刪除

使用兩個(gè)隊(duì)列實(shí)現(xiàn)堆棧 (Implementation of a stack using two queues)

Likewise, a queue can be implemented with two stacks, a stack can also be implemented using two queues. The basic idea is to perform stack ADT operations using the two queues.

同樣，一個(gè)隊(duì)列可以用兩個(gè)堆棧實(shí)現(xiàn)，一個(gè)堆棧也可以用兩個(gè)隊(duì)列實(shí)現(xiàn)。 基本思想是使用兩個(gè)隊(duì)列執(zhí)行堆棧ADT操作。

So, we need to implement push(),pop() using DeQueue(), EnQueue() operations available for the queues.

因此，我們需要使用隊(duì)列可用的DeQueue()和EnQueue()操作來(lái)實(shí)現(xiàn)push()，pop()。

Implementation:

實(shí)現(xiàn)方式：

Let q1 and q2 be the two queues...

設(shè)q1和q2為兩個(gè)隊(duì)列...

struct stack{struct queue *q1;struct queue *q2; }

Algorithm to implement push and pop:

實(shí)現(xiàn)推送和彈出的算法：

Push operation algorithm:

推送操作算法：

Check whether q1 is empty or not. If q1 is empty then EnQueue the element to q2.

檢查q1是否為空。 如果q1為空，則將元素排隊(duì)到q2。

Otherwise EnQueue to q1.

否則，排隊(duì)到q1。

push(struct stack *s,int data){if(isempty(s->q1))EnQueue(s->q2,data);elseEnQueue(s->q1,data); }

Pop operation algorithm

彈出操作算法

The basic idea is to transfer n-1 elements (let n be the total no of elements) to other queue and delete the last one from a queue to perform the pop operation.

基本思想是將n-1個(gè)元素(使n為元素總數(shù))轉(zhuǎn)移到其他隊(duì)列，并從隊(duì)列中刪除最后一個(gè)元素以執(zhí)行彈出操作。

If q1 is not empty then transfer n-1 elements from q1 to q2 and DeQueue the last element and return it.

如果q1不為空，則將n-1個(gè)元素從q1傳輸?shù)絨2，并對(duì)最后一個(gè)元素進(jìn)行DeQueue并返回。

If q2 is not empty then transfer n-1 elements from q2 to q1 and DeQueue the last element and return it.

如果q2不為空，則將n-1個(gè)元素從q2傳輸?shù)絨1，并對(duì)最后一個(gè)元素進(jìn)行DeQueue并返回。

int pop(struct stack *s){int count,size;if(isempty(s->q2)){size=Size(s->q1);count=0;while(count<size-1){//transferring n-1 elementsEnQueue(s->q2,DeQueue(s->q1));count++;}//last element to be poppedreturn DeQueue(s->q1);}else{size=Size(s->q2);count=0;while(count<size-1){EnQueue(s->q1,DeQueue(s->q2));count++;}return DeQueue(s->q1);} }

Time complexity analysis:

時(shí)間復(fù)雜度分析：

Push: O(1)

推：O(1)

Pop: O(n) , since transferring n-1 elements

彈出：O(n)，因?yàn)閭鬏斄薾-1個(gè)元素

.minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

使用C ++的實(shí)現(xiàn)代碼(使用STL) (Implementation code using C++ (using STL))

#include <bits/stdc++.h>using namespace std;struct Stack{queue<int> q1,q2;void push(int x){if(q1.empty()){q2.push(x); //EnQueue operation using STL}else{q1.push(x); //EnQueue operation using STL}}int pop(){int count,size,item;if(q2.empty()){size=q1.size(); //size=no of elements;count=0;while(count<size-1){ //transfering n-1 elementsq2.push(q1.front()); // DeQueue operation using STLq1.pop();count++;}item=q1.front();q1.pop();return item; //popping out the element}else{size=q2.size();count=0;while(count<size-1){q1.push(q2.front());q2.pop();count++;}item=q2.front();q2.pop();return item;}} };int main() {cout<<"implementing stack with two queues"<<endl;cout<<"enter any integer to push and 0 to stop pushing"<<endl;Stack s;int x,count=0;cin>>x;while(x){s.push(x);cin>>x;count++;}cout<<"now popping......."<<endl;while(count){cout<<s.pop()<<endl;count--;}cout<<"executed successfully!!!"<<endl;return 0; }

Output

輸出量

implementing stack with two queues enter any integer to push and 0 to stop pushing 1 2 3 0 now popping....... 3 2 1 executed successfully!!!

翻譯自: https://www.includehelp.com/data-structure-tutorial/implementation-of-stack-using-two-queues.aspx

如何使用兩個(gè)堆棧實(shí)現(xiàn)隊(duì)列

總結(jié)

以上是生活随笔為你收集整理的如何使用两个堆栈实现队列_使用两个队列实现堆栈的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。