Reverse a stack without using extra space in O(n)
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++
// C++ program to implement Stack // using linked list so that reverse // can be done with O(1) extra space. #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 : // Push and pop operations 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; } // prints contents of stack void display() { StackNode *s = top; while (s != NULL) { cout << s->data << " " ; s = s->next; } cout << endl; } // Reverses the stack using simple // linked list reversal logic. 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; } }; // driver code 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; // reverse s->reverse(); cout << "Reversed Stack" << endl; s->display(); return 0; } // This code is contributed by Chhavi. |
Java
// Java program to implement Stack using linked // list so that reverse can be done with O(1) // extra space. class StackNode { int data; StackNode next; public StackNode( int data) { this .data = data; this .next = null ; } } class Stack { StackNode top; // Push and pop operations 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; } // prints contents of stack public void display() { StackNode s = this .top; while (s != null ) { System.out.print(s.data + " " ); s = s.next; } System.out.println(); } // Reverses the stack using simple // linked list reversal logic. 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(); // reverse s.reverse(); System.out.println( "Reversed Stack" ); s.display(); } } |
Python3
# Python3 program to implement Stack # using linked list so that reverse # can be done with O(1) extra space. class StackNode: def __init__( self , data): self .data = data self . next = None class Stack: def __init__( self ): self .top = None # Push and pop operations 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 # Prints contents of stack def display( self ): s = self .top while (s ! = None ): print (s.data, end = ' ' ) s = s. next # Reverses the stack using simple # linked list reversal logic. 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 # Driver code if __name__ = = '__main__' : s = Stack() s.push( 1 ) s.push( 2 ) s.push( 3 ) s.push( 4 ) print ( "Original Stack" ) s.display() print () # Reverse s.reverse() print ( "Reversed Stack" ) s.display() # This code is contributed by rutvik_56 |
C#
// C# program to implement Stack using linked // list so that reverse can be done with O(1) // extra space. 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; // Push and pop operations 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; } // prints contents of stack public void display() { StackNode s = this .top; while (s != null ) { Console.Write(s.data + " " ); s = s.next; } Console.WriteLine(); } // Reverses the stack using simple // linked list reversal logic. 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 { // Driver code 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(); // reverse s.reverse(); Console.WriteLine( "Reversed Stack" ); s.display(); } } // This code is contributed by Arnab Kundu |
Javascript
<script> // JavaScript program to implement Stack // using linked list so that reverse can // be done with O(1) extra space. class StackNode { constructor(data) { this .data = data; this .next = null ; } } class Stack { top = null ; // Push and pop operations 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; } // Prints contents of stack display() { var s = this .top; while (s != null ) { document.write(s.data + " " ); s = s.next; } document.write( "<br>" ); } // Reverses the stack using simple // linked list reversal logic. 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; } } // Driver code var s = new Stack(); s.push(1); s.push(2); s.push(3); s.push(4); document.write( "Original Stack <br>" ); s.display(); // Reverse s.reverse(); document.write( "Reversed Stack <br>" ); s.display(); // This code is contributed by rdtank </script> |
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.
This article is contributed by Niharika Sahai. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Login to comment...