Reverse a Stack without using recursion and extra space. Even the functional Stack is not allowed.
Examples:
Input : 1->2->3->4
Output : 4->3->2->1
Input : 6->5->4
Output : 4->5->6
We have discussed a way of reversing a stack in the below post.
Reverse a Stack using Recursion
The above solution requires O(n) extra space. We can reverse a stack in O(1) time if we internally represent the stack as a linked list. Reverse a stack would require a reversing of a linked list which can be done with O(n) time and O(1) extra space.
Note that push() and pop() operations still take O(1) time.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
class StackNode {
public :
int data;
StackNode *next;
StackNode( int data)
{
this ->data = data;
this ->next = NULL;
}
};
class Stack {
StackNode *top;
public :
void push( int data)
{
if (top == NULL) {
top = new StackNode(data);
return ;
}
StackNode *s = new StackNode(data);
s->next = top;
top = s;
}
StackNode* pop()
{
StackNode *s = top;
top = top->next;
return s;
}
void display()
{
StackNode *s = top;
while (s != NULL) {
cout << s->data << " " ;
s = s->next;
}
cout << endl;
}
void reverse()
{
StackNode *prev, *cur, *succ;
cur = prev = top;
cur = cur->next;
prev->next = NULL;
while (cur != NULL) {
succ = cur->next;
cur->next = prev;
prev = cur;
cur = succ;
}
top = prev;
}
};
int main()
{
Stack *s = new Stack();
s->push(1);
s->push(2);
s->push(3);
s->push(4);
cout << "Original Stack" << endl;;
s->display();
cout << endl;
s->reverse();
cout << "Reversed Stack" << endl;
s->display();
return 0;
}
|
Java
class StackNode {
int data;
StackNode next;
public StackNode( int data)
{
this .data = data;
this .next = null ;
}
}
class Stack {
StackNode top;
public void push( int data)
{
if ( this .top == null ) {
top = new StackNode(data);
return ;
}
StackNode s = new StackNode(data);
s.next = this .top;
this .top = s;
}
public StackNode pop()
{
StackNode s = this .top;
this .top = this .top.next;
return s;
}
public void display()
{
StackNode s = this .top;
while (s != null ) {
System.out.print(s.data + " " );
s = s.next;
}
System.out.println();
}
public void reverse()
{
StackNode prev, cur, succ;
cur = prev = this .top;
cur = cur.next;
prev.next = null ;
while (cur != null ) {
succ = cur.next;
cur.next = prev;
prev = cur;
cur = succ;
}
this .top = prev;
}
}
public class reverseStackWithoutSpace {
public static void main(String[] args)
{
Stack s = new Stack();
s.push( 1 );
s.push( 2 );
s.push( 3 );
s.push( 4 );
System.out.println( "Original Stack" );
s.display();
s.reverse();
System.out.println( "Reversed Stack" );
s.display();
}
}
|
Python3
class StackNode:
def __init__( self , data):
self .data = data
self . next = None
class Stack:
def __init__( self ):
self .top = None
def push( self , data):
if ( self .top = = None ):
self .top = StackNode(data)
return
s = StackNode(data)
s. next = self .top
self .top = s
def pop( self ):
s = self .top
self .top = self .top. next
return s
def display( self ):
s = self .top
while (s ! = None ):
print (s.data, end = ' ' )
s = s. next
def reverse( self ):
prev = self .top
cur = self .top
cur = cur. next
succ = None
prev. next = None
while (cur ! = None ):
succ = cur. next
cur. next = prev
prev = cur
cur = succ
self .top = prev
if __name__ = = '__main__' :
s = Stack()
s.push( 1 )
s.push( 2 )
s.push( 3 )
s.push( 4 )
print ( "Original Stack" )
s.display()
print ()
s.reverse()
print ( "Reversed Stack" )
s.display()
|
C#
using System;
public class StackNode
{
public int data;
public StackNode next;
public StackNode( int data)
{
this .data = data;
this .next = null ;
}
}
public class Stack
{
public StackNode top;
public void push( int data)
{
if ( this .top == null )
{
top = new StackNode(data);
return ;
}
StackNode s = new StackNode(data);
s.next = this .top;
this .top = s;
}
public StackNode pop()
{
StackNode s = this .top;
this .top = this .top.next;
return s;
}
public void display()
{
StackNode s = this .top;
while (s != null )
{
Console.Write(s.data + " " );
s = s.next;
}
Console.WriteLine();
}
public void reverse()
{
StackNode prev, cur, succ;
cur = prev = this .top;
cur = cur.next;
prev.next = null ;
while (cur != null )
{
succ = cur.next;
cur.next = prev;
prev = cur;
cur = succ;
}
this .top = prev;
}
}
public class reverseStackWithoutSpace
{
public static void Main(String []args)
{
Stack s = new Stack();
s.push(1);
s.push(2);
s.push(3);
s.push(4);
Console.WriteLine( "Original Stack" );
s.display();
s.reverse();
Console.WriteLine( "Reversed Stack" );
s.display();
}
}
|
Javascript
<script>
class StackNode
{
constructor(data)
{
this .data = data;
this .next = null ;
}
}
class Stack
{
top = null ;
push(data)
{
if ( this .top == null )
{
this .top = new StackNode(data);
return ;
}
var s = new StackNode(data);
s.next = this .top;
this .top = s;
}
pop()
{
var s = this .top;
this .top = this .top.next;
return s;
}
display()
{
var s = this .top;
while (s != null )
{
document.write(s.data + " " );
s = s.next;
}
document.write( "<br>" );
}
reverse()
{
var prev, cur, succ;
cur = prev = this .top;
cur = cur.next;
prev.next = null ;
while (cur != null )
{
succ = cur.next;
cur.next = prev;
prev = cur;
cur = succ;
}
this .top = prev;
}
}
var s = new Stack();
s.push(1);
s.push(2);
s.push(3);
s.push(4);
document.write( "Original Stack <br>" );
s.display();
s.reverse();
document.write( "Reversed Stack <br>" );
s.display();
</script>
|
Output
Original Stack
4 3 2 1
Reversed Stack
1 2 3 4
Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.
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