# Java Program For Rearranging A Given Linked List In-Place.

• Last Updated : 15 Jun, 2022

Given a singly linked list L0 -> L1 -> … -> Ln-1 -> Ln. Rearrange the nodes in the list so that the new formed list is : L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2
You are required to do this in place without altering the nodes’ values.

Examples:

```Input: 1 -> 2 -> 3 -> 4
Output: 1 -> 4 -> 2 -> 3

Input: 1 -> 2 -> 3 -> 4 -> 5
Output: 1 -> 5 -> 2 -> 4 -> 3```

Simple Solution:

```1) Initialize current node as head.
2) While next of current node is not null, do following
a) Find the last node, remove it from the end and insert it as next
of the current node.
b) Move current to next to next of current```

The time complexity of the above simple solution is O(n2) where n is the number of nodes in the linked list.

Better Solution:
1) Copy contents of the given linked list to a vector.
2) Rearrange the given vector by swapping nodes from both ends.
3) Copy the modified vector back to the linked list.
Implementation of this approach: https://ide.geeksforgeeks.org/1eGSEy
Thanks to Arushi Dhamija for suggesting this approach.

Efficient Solution:

```1) Find the middle point using tortoise and hare method.
2) Split the linked list into two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.```

The Time Complexity of this solution is O(n).

Below is the implementation of this method.

## Java

 `// Java program to rearrange linked list ` `// in place`   `// Linked List Class` `class` `LinkedList ` `{` `    ``// head of the list` `    ``static` `Node head; `   `    ``// Node Class ` `    ``static` `class` `Node ` `    ``{` `        ``int` `data;` `        ``Node next;`   `        ``// Constructor to create ` `        ``// a new node` `        ``Node(``int` `d)` `        ``{` `            ``data = d;` `            ``next = ``null``;` `        ``}` `    ``}`   `    ``void` `printlist(Node node)` `    ``{` `        ``if` `(node == ``null``) ` `        ``{` `            ``return``;` `        ``}` `        ``while` `(node != ``null``) ` `        ``{` `            ``System.out.print(node.data + ` `                             ``" -> "``);` `            ``node = node.next;` `        ``}` `    ``}`   `    ``Node reverselist(Node node)` `    ``{` `        ``Node prev = ``null``, ` `             ``curr = node, next;` `        ``while` `(curr != ``null``) ` `        ``{` `            ``next = curr.next;` `            ``curr.next = prev;` `            ``prev = curr;` `            ``curr = next;` `        ``}` `        ``node = prev;` `        ``return` `node;` `    ``}`   `    ``void` `rearrange(Node node)` `    ``{` `        ``// 1) Find the middle point using ` `        ``// tortoise and hare method` `        ``Node slow = node, fast = slow.next;` `        ``while` `(fast != ``null` `&& ` `               ``fast.next != ``null``) ` `        ``{` `            ``slow = slow.next;` `            ``fast = fast.next.next;` `        ``}`   `        ``// 2) Split the linked list in ` `        ``// two halves` `        ``// node1, head of first half- ` `        ``// 1 -> 2 -> 3` `        ``// node2, head of second half- ` `        ``// 4 -> 5` `        ``Node node1 = node;` `        ``Node node2 = slow.next;` `        ``slow.next = ``null``;`   `        ``// 3) Reverse the second half, ` `        ``// i.e., 5 -> 4` `        ``node2 = reverselist(node2);`   `        ``// 4) Merge alternate nodes` `        ``// Assign dummy Node` `        ``node = ``new` `Node(``0``); `   `        ``// curr is the pointer to this ` `        ``// dummy Node, which will be ` `        ``// used to form the new list` `        ``Node curr = node;` `        ``while` `(node1 != ``null` `|| ` `               ``node2 != ``null``) ` `        ``{` `            ``// First add the element ` `            ``// from first list` `            ``if` `(node1 != ``null``) ` `            ``{` `                ``curr.next = node1;` `                ``curr = curr.next;` `                ``node1 = node1.next;` `            ``}`   `            ``// Then add the element from ` `            ``// second list` `            ``if` `(node2 != ``null``) ` `            ``{` `                ``curr.next = node2;` `                ``curr = curr.next;` `                ``node2 = node2.next;` `            ``}` `        ``}`   `        ``// Assign the head of the new ` `        ``// list to head pointer` `        ``node = node.next;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``LinkedList list = ``new` `LinkedList();` `        ``list.head = ``new` `Node(``1``);` `        ``list.head.next = ``new` `Node(``2``);` `        ``list.head.next.next = ` `        ``new` `Node(``3``);` `        ``list.head.next.next.next = ` `        ``new` `Node(``4``);` `        ``list.head.next.next.next.next = ` `        ``new` `Node(``5``);` ` `  `        ``// Print original list` `        ``list.printlist(head); `   `        ``// Rearrange list as per ques` `        ``list.rearrange(head); ` `        ``System.out.println(``""``);`   `        ``// Print modified list` `        ``list.printlist(head); ` `    ``}` `}` `// This code is contributed by Mayank Jaiswal`

Output:

```1 -> 2 -> 3 -> 4 -> 5
1 -> 5 -> 2 -> 4 -> 3```

Time Complexity: O(n)
Auxiliary Space: O(1)
Thanks to Gaurav Ahirwar for suggesting the above approach.

Another approach:
2. Compare their data and swap.
After that, a new linked list is formed.

Below is the implementation:

## Java

 `// Java code to rearrange linked list ` `// in place` `class` `Geeks ` `{` `    ``static` `class` `Node ` `    ``{` `        ``int` `data;` `        ``Node next;` `    ``}`   `    ``// Function for rearranging a ` `    ``// linked list with high and ` `    ``// low value.` `    ``static` `Node rearrange(Node head)` `    ``{` `        ``// Base case` `        ``if` `(head == ``null``) ` `            ``return` `null``;`   `        ``// Two pointer variable.` `        ``Node prev = head, ` `             ``curr = head.next;`   `        ``while` `(curr != ``null``) ` `        ``{` `            ``// Swap function for swapping ` `            ``// data.` `            ``if` `(prev.data > curr.data) ` `            ``{` `                ``int` `t = prev.data;` `                ``prev.data = curr.data;` `                ``curr.data = t;` `            ``}`   `            ``// Swap function for swapping data.` `            ``if` `(curr.next != ``null` `&& ` `                ``curr.next.data > curr.data) ` `            ``{` `                ``int` `t = curr.next.data;` `                ``curr.next.data = curr.data;` `                ``curr.data = t;` `            ``}`   `            ``prev = curr.next;`   `            ``if` `(curr.next == ``null``)` `                ``break``;` `            ``curr = curr.next.next;` `        ``}` `        ``return` `head;` `    ``}`   `    ``// Function to insert a Node in` `    ``// the linked list at the beginning.` `    ``static` `Node push(Node head, ``int` `k)` `    ``{` `        ``Node tem = ``new` `Node();` `        ``tem.data = k;` `        ``tem.next = head;` `        ``head = tem;` `        ``return` `head;` `    ``}`   `    ``// Function to display Node of ` `    ``// linked list.` `    ``static` `void` `display(Node head)` `    ``{` `        ``Node curr = head;` `        ``while` `(curr != ``null``) ` `        ``{` `            ``System.out.printf(``"%d "``,` `                              ``curr.data);` `            ``curr = curr.next;` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``Node head = ``null``;`   `        ``// Let create a linked list.` `        ``// 9 . 6 . 8 . 3 . 7` `        ``head = push(head, ``7``);` `        ``head = push(head, ``3``);` `        ``head = push(head, ``8``);` `        ``head = push(head, ``6``);` `        ``head = push(head, ``9``);` `        ``head = rearrange(head);` `        ``display(head);` `    ``}` `}` `// This code is contributed by Arnab Kundu`

Output:

`6 9 3 8 7`

Time Complexity : O(n)
Auxiliary Space : O(1)
Thanks to Aditya for suggesting this approach.

Another Approach: (Using recursion)

1. Hold a pointer to the head node and go till the last node using recursion
2. Once the last node is reached, start swapping the last node to the next of head node
3. Move the head pointer to the next node
4. Repeat this until the head and the last node meet or come adjacent to each other
5. Once the Stop condition met, we need to discard the left nodes to fix the loop created in the list while swapping nodes.

## Java

 `// Java implementation` `import` `java.io.*;` `// Java program to implement` `// the above approach` `// Creating the structure ` `// for node` `class` `Node ` `{` `    ``int` `data;` `    ``Node next;`   `    ``// Function to create newNode ` `    ``// in a linkedlist` `    ``Node(``int` `key)` `    ``{` `        ``data = key;` `        ``next = ``null``;` `    ``}` `}` `class` `GFG ` `{` `    ``Node left = ``null``;`   `    ``// Function to print the list` `    ``void` `printlist(Node head)` `    ``{` `        ``while` `(head != ``null``) ` `        ``{` `            ``System.out.print(head.data + ` `                             ``" "``);` `            ``if` `(head.next != ``null``) ` `            ``{` `                ``System.out.print(``"->"``);` `            ``}` `            ``head = head.next;` `        ``}` `        ``System.out.println();` `    ``}`   `    ``// Function to rearrange` `    ``void` `rearrange(Node head)` `    ``{` `        ``if` `(head != ``null``) ` `        ``{` `            ``left = head;` `            ``reorderListUtil(left);` `        ``}` `    ``}`   `    ``void` `reorderListUtil(Node right)` `    ``{` `        ``if` `(right == ``null``) ` `        ``{` `            ``return``;` `        ``}`   `        ``reorderListUtil(right.next);`   `        ``// We set left = null, when we ` `        ``// reach stop condition, so no ` `        ``// processing required after that` `        ``if` `(left == ``null``) ` `        ``{` `            ``return``;` `        ``}`   `        ``// Stop condition: odd case : ` `        ``// left = right, even` `        ``// case : left.next = right` `        ``if` `(left != right && ` `            ``left.next != right) ` `        ``{` `            ``Node temp = left.next;` `            ``left.next = right;` `            ``right.next = temp;` `            ``left = temp;` `        ``}` `        ``else` `        ``{ ` `            ``// stop condition , set null ` `            ``// to left nodes` `            ``if` `(left.next == right) ` `            ``{` `                ``// even case` `                ``left.next.next = ``null``; ` `                ``left = ``null``;` `            ``}` `            ``else` `            ``{` `                ``// odd case` `                ``left.next = ``null``; ` `                ``left = ``null``;` `            ``}` `        ``}` `    ``}`   `    ``// Drivers Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``Node head = ``new` `Node(``1``);` `        ``head.next = ``new` `Node(``2``);` `        ``head.next.next = ``new` `Node(``3``);` `        ``head.next.next.next = ` `        ``new` `Node(``4``);` `        ``head.next.next.next.next = ` `        ``new` `Node(``5``);` `      `  `        ``GFG gfg = ``new` `GFG();`   `        ``// Print original list` `        ``gfg.printlist(head);        `   `        ``// Modify the list` `        ``gfg.rearrange(head);`   `        ``// Print modified list` `        ``gfg.printlist(head);` `    ``}` `}` `// This code is contributed by Vishal Singh`

Output:

```1 ->2 ->3 ->4 ->5
1 ->5 ->2 ->4 ->3```

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive stack where n represents the length of the given linked list.

Please refer complete article on Rearrange a given linked list in-place. for more details!

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