Linked List - Swap nodes

Swap nodes in a linked list without swapping the data

 Sample input: swap 2 and 9 in 1->5->2->3->9->4
Sample output: 1->5->9->3->2->4

 Steps:
1) find X,Y in the list and also their previous nodes.
2) prevX.next = Y and prevY.next = X
3) temp = X.next , X.next = Y.next, Y.next = temp

 Code:

 void swapNodes(int x, int y){

         Node prevX = null, X = head;
        while (X != null && X.data != x)
        {
            prevX = X;
            X = X.next;
        }

         Node prevY = null, Y = head;
        while (Y != null && Y.data != y)
        {
            prevY = Y;
            Y = Y.next;
        }

         if (prevX != null)
            prevX.next = Y;
        else 
            head = Y;

         if (prevY != null)
            prevY.next = X;
        else 
            head = X;

         Node temp = X.next;
        X.next = Y.next;
        Y.next = temp;
    }
Illustration:

 Example (i):
swap 2 and 9 in 1->5->2->3->9->4

 Find prevX,X,prevY,Y

prevX.next = Y

prevY.next = X
temp = X.next
X.next = Y.next
X.next = temp

Result:

Example (ii): (one of the node is head node)

 Swap 1 and 2 in 1->2->3->4->5

 Find X and Y and their prev nodes

Since prevX = null, head=Y

prevY.next = X
swap next nodes of x and y

                  Happy New Year :)






Comments

Post a Comment

Popular posts from this blog

Tree traversals - Inorder Traversal

Image
        A binary tree can be traversed in two ways. i) Depth First Traversal: Inorder, preorder, postorder ii) Breadth First Traversal: Level Order Traversal Inorder traversal:        In inorder traversal root is visited in between left and right subtree traversals. Steps: 1.  Traverse the left subtree 2.  Visit the root 3. Traverse the right subtree Note: If binary search tree is traversed in inorder, the numbers will be sorted in ascending order. Code: void inOrder(Node root) { if(root==null) return; inOrder(root.left); System.out.print(root.data+" "); inOrder(root.right); } Illustration: Inorder traversal of given tree is 3, 5, 4, 7, 8  Consider the left subtree of 7 (since 7 is the root ,we always start our traversal from root) Here we start form 5, the left subtree of 5 is 3. For 3 the left subtree is empty , So we visit the root. The right subtree of tree is also empty. At ...
Read more

Binary Search tree (Insertion)

Image
   A tree in which every node can have a maximum of two children is called a binary tree.       A binary search tree is a binary tree in which every node has lower values in the left sub tree and higher values in the right sub tree. Node of binary tree: class Node{      int data;      Node left, right;      Node(int data){         this.data = data;         right = left = null;      } } BST - Insertion:  static Node Insert(Node root,int data) {      if( root == null ){            root = new Node(data);            return root;      }      if( data < root.data)             root.left = Insert(root.left,data);      else             root.right = Insert(root.rig...
Read more

Stack implementation using array

Image
    Introduction:    Stack is a linear data structure which follows LIFO( Last In First Out ) order.   To represent stack consider books are placed in table for the teacher to correct. The note which is submitted last will be corrected first. The person who submits last is quite unlucky isn't it :P Operations: push: Inserts element at the top of the stack pop: Removes element from the top of the stack Stack implementation using array: Code: import java.util.*; import java.lang.*; import java.io.*; class Stack {     int SIZE = 10;     int arr[] = new int[SIZE];     int top = -1;          public int pop(){         if (top == -1){             System.out.println("stack is empty");             return 0;         }         int topelement ...
Read more