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 string in below post.
Reverse a Stack using Recursion
The above solution requires O(n) extra space. We can reverse a string in O(1) time if we internally represent the stack as a linked list. Reverse a stack would require a reversing 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.
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 codeint 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 contribute 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 = Noneclass 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 codeif __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 |
Output:
Original Stack 4 3 2 1 Reversed Stack 1 2 3 4
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