原文：
给定s，任务是，以使元素仍然存在于栈中，而不改变它们的顺序。
示例：
输入：
s = {2, 3, 4, 5}输出：
5 4 3 2输入：
s = {3, 3, 2, 2}输出：
2 2 3 3
：请按照以下步骤解决问题：
-
创建一个具有作为参数的递归函数。
-
添加基本条件，如果栈为空，则从函数返回。
-
否则，将顶部元素存储在某个变量
x中，然后将其删除。 -
打印
x，调用递归函数并在其中传递相同的栈。 -
将存储的
x推回栈。
下面是上述方法的实现：
c
// c program for the above approach
#include
using namespace std;
// function to print stack elements
// from top to bottom with the
// order of elements unaltered
void printstack(stack s)
{
// if stack is empty
if (s.empty())
return;
// extract top of the stack
int x = s.top();
// pop the top element
s.pop();
// print the current top
// of the stack i.e., x
cout << x << ' ';
// proceed to print
// remaining stack
printstack(s);
// push the element back
s.push(x);
}
// driver code
int main()
{
stack s;
// given stack s
s.push(1);
s.push(2);
s.push(3);
s.push(4);
// function call
printstack(s);
return 0;
}
java
// java program for the above approach
import java.util.*;
class gfg{
// function to print stack elements
// from top to bottom with the
// order of elements unaltered
public static void printstack(stack s)
{
// if stack is empty
if (s.empty())
return;
// extract top of the stack
int x = s.peek();
// pop the top element
s.pop();
// print the current top
// of the stack i.e., x
system.out.print(x " ");
// proceed to print
// remaining stack
printstack(s);
// push the element back
s.push(x);
}
// driver code
public static void main(string[] args)
{
stack s = new stack();
// given stack s
s.push(1);
s.push(2);
s.push(3);
s.push(4);
// function call
printstack(s);
}
}
// this code is contributed divyeshrabadiya07
python3
# python3 program for the
# above approach
from queue import lifoqueue
# function to prstack elements
# from top to bottom with the
# order of elements unaltered
def printstack(s):
# if stack is empty
if (s.empty()):
return;
# extract top of the
# stack
x = s.get();
# pop the top element
#s.pop();
# print current top
# of the stack i.e., x
print(x, end = " ");
# proceed to print
# remaining stack
printstack(s);
# push the element
# back
s.put(x);
# driver code
if __name__ == '__main__':
s = lifoqueue();
# given stack s
s.put(1);
s.put(2);
s.put(3);
s.put(4);
# function call
printstack(s);
# this code is contributed by amit katiyar
c
// c# program for
// the above approach
using system;
using system.collections.generic;
class gfg{
// function to print stack elements
// from top to bottom with the
// order of elements unaltered
public static void printstack(stack s)
{
// if stack is empty
if (s.count == 0)
return;
// extract top of the stack
int x = s.peek();
// pop the top element
s.pop();
// print the current top
// of the stack i.e., x
console.write(x " ");
// proceed to print
// remaining stack
printstack(s);
// push the element back
s.push(x);
}
// driver code
public static void main(string[] args)
{
stack s = new stack();
// given stack s
s.push(1);
s.push(2);
s.push(3);
s.push(4);
// function call
printstack(s);
}
}
// this code is contributed by rajput-ji
输出：
4 3 2 1
时间复杂度：o(n)，其中n是给定栈中元素的数量。
辅助空间：o(n)
：此方法讨论解决表示法的问题的pg电子试玩链接的解决方案。 步骤如下：
-
将给定栈中的顶部元素推入。
-
打印的顶部元素。
-
从给定的栈中弹出顶部元素。
-
按顺序重复上述步骤，直到给定的栈为空。
下面是上述方法的实现：
java
// java program for the above approach
import static java.lang.system.exit;
// create stack using linked list
class stackusinglinkedlist {
// a linked list node
private class node {
int data;
node link;
}
// reference variable
node top;
// constructor
stackusinglinkedlist()
{
this.top = null;
}
// function to add an element x
// in the stack by inserting
// at the beginning of ll
public void push(int x)
{
// create new node temp
node temp = new node();
// if stack is full
if (temp == null) {
system.out.print("\nheap overflow");
return;
}
// initialize data into temp
temp.data = x;
// reference into temp link
temp.link = top;
// update top reference
top = temp;
}
// function to check if the stack
// is empty or not
public boolean isempty()
{
return top == null;
}
// function to return top element
// in a stack
public int peek()
{
// check for empty stack
if (!isempty()) {
// return the top data
return top.data;
}
// otherwise stack is empty
else {
system.out.println("stack is empty");
return -1;
}
}
// function to pop top element from
// the stack by removing element
// at the beginning
public void pop()
{
// check for stack underflow
if (top == null) {
system.out.print("\nstack underflow");
return;
}
// update the top pointer to
// point to the next node
top = (top).link;
}
// function to print the elements and
// restore the stack
public static void
displaystack(stackusinglinkedlist s)
{
// create another stack
stackusinglinkedlist s1
= new stackusinglinkedlist();
// until stack is empty
while (!s.isempty()) {
s1.push(s.peek());
// print the element
system.out.print(s1.peek()
" ");
s.pop();
}
}
}
// driver code
public class gfg {
// driver code
public static void main(string[] args)
{
// create object of class
stackusinglinkedlist obj
= new stackusinglinkedlist();
// insert stack value
obj.push(1);
obj.push(2);
obj.push(3);
obj.push(4);
// function call
obj.displaystack(obj);
}
}
c
// c# program for the above approach
using system;
// create stack using linked list
class stackusinglinkedlist{
// a linked list node
public class node
{
public int data;
public node link;
}
// reference variable
node top;
// constructor
public stackusinglinkedlist()
{
this.top = null;
}
// function to add an element x
// in the stack by inserting
// at the beginning of ll
public void push(int x)
{
// create new node temp
node temp = new node();
// if stack is full
if (temp == null)
{
console.write("\nheap overflow");
return;
}
// initialize data into temp
temp.data = x;
// reference into temp link
temp.link = top;
// update top reference
top = temp;
}
// function to check if the stack
// is empty or not
public bool isempty()
{
return top == null;
}
// function to return top element
// in a stack
public int peek()
{
// check for empty stack
if (isempty() != true)
{
// return the top data
return top.data;
}
// otherwise stack is empty
else
{
console.writeline("stack is empty");
return -1;
}
}
// function to pop top element from
// the stack by removing element
// at the beginning
public void pop()
{
// check for stack underflow
if (top == null)
{
console.write("\nstack underflow");
return;
}
// update the top pointer to
// point to the next node
top = (top).link;
}
// function to print the elements and
// restore the stack
public void displaystack(stackusinglinkedlist s)
{
// create another stack
stackusinglinkedlist s1 = new stackusinglinkedlist();
// until stack is empty
while (s.isempty() != true)
{
s1.push(s.peek());
// print the element
console.write(s1.peek() " ");
s.pop();
}
}
}
class gfg{
// driver code
public static void main(string[] args)
{
// create object of class
stackusinglinkedlist obj = new stackusinglinkedlist();
// insert stack value
obj.push(1);
obj.push(2);
obj.push(3);
obj.push(4);
// function call
obj.displaystack(obj);
}
}
// this code is contributed by amit katiyar
输出：
4 3 2 1
时间复杂度：o(n)，其中n是给定栈中元素的数量。
辅助空间：o(n)
：此方法讨论实现中问题的pg电子试玩链接的解决方案。 步骤如下：
-
将给定栈中的顶部元素推入。
-
打印的顶部元素。
-
从给定的栈中弹出顶部元素。
-
按顺序重复上述步骤，直到给定的栈为空。
下面是上述方法的实现：
java
// java program for the above approach
class stack {
static final int max = 1000;
// stores the index where element
// needs to be inserted
int top;
// array to store the stack elements
int a[] = new int[max];
// function that check whether stack
// is empty or not
boolean isempty()
{
return (top < 0);
}
// constructor
stack() { top = -1; }
// function that pushes the element
// to the top of the stack
boolean push(int x)
{
// if stack is full
if (top >= (max - 1)) {
system.out.println(
"stack overflow");
return false;
}
// otherwise insert element x
else {
a[ top] = x;
return true;
}
}
// function that removes the top
// element from the stack
int pop()
{
// if stack is empty
if (top < 0) {
system.out.println(
"stack underflow");
return 0;
}
// otherwise remove element
else {
int x = a[top--];
return x;
}
}
// function to get the top element
int peek()
{
// if stack is empty
if (top < 0) {
system.out.println(
"stack underflow");
return 0;
}
// otherwise remove element
else {
int x = a[top];
return x;
}
}
// function to print the elements
// and restore the stack
static void displaystack(stack s)
{
// create another stack
stack s1 = new stack();
// until stack is empty
while (!s.isempty()) {
s1.push(s.peek());
// print the element
system.out.print(s1.peek()
" ");
s.pop();
}
}
}
// driver code
class main {
// driver code
public static void main(string args[])
{
stack s = new stack();
// given stack
s.push(1);
s.push(2);
s.push(3);
s.push(4);
// function call
s.displaystack(s);
}
}
c
// c# program for
// the above approach
using system;
class stack{
static int max = 1000;
// stores the index where
// element needs to be inserted
int top;
// array to store the
// stack elements
int []a = new int[max];
// function that check
// whether stack
// is empty or not
bool isempty()
{
return (top < 0);
}
// constructor
public stack()
{
top = -1;
}
// function that pushes
// the element to the
// top of the stack
public bool push(int x)
{
// if stack is full
if (top >= (max - 1))
{
console.writeline("stack overflow");
return false;
}
// otherwise insert element x
else
{
a[ top] = x;
return true;
}
}
// function that removes the top
// element from the stack
public int pop()
{
// if stack is empty
if (top < 0)
{
console.writeline("stack underflow");
return 0;
}
// otherwise remove element
else
{
int x = a[top--];
return x;
}
}
// function to get the top element
public int peek()
{
// if stack is empty
if (top < 0)
{
console.writeline("stack underflow");
return 0;
}
// otherwise remove element
else
{
int x = a[top];
return x;
}
}
// function to print the elements
// and restore the stack
public void displaystack(stack s)
{
// create another stack
stack s1 = new stack();
// until stack is empty
while (!s.isempty())
{
s1.push(s.peek());
// print the element
console.write(s1.peek() " ");
s.pop();
}
}
}
class gfg{
// driver code
public static void main(string []args)
{
stack s = new stack();
// given stack
s.push(1);
s.push(2);
s.push(3);
s.push(4);
// function call
s.displaystack(s);
}
}
// this code is contributed by 29ajaykumar
输出：
4 3 2 1
时间复杂度：o(n)，其中n是给定栈中元素的数量。
辅助空间：o(n)
如果您喜欢 geeksforgeeks 并希望做出贡献，则还可以使用 tribution.geeksforgeeks.org 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。 查看您的文章出现在 geeksforgeeks pg电子试玩链接主页上，并帮助其他 geeks。
如果您发现任何不正确的地方，请单击下面的“改进文章”按钮，以改进本文。
麻将胡了pg电子网站的版权属于：月萌api www.moonapi.com，转载请注明出处
