# Python Program For Swapping Nodes In A Linked List Without Swapping Data

Given a linked list and two keys in it, swap nodes for two given keys. Nodes should be swapped by changing links. Swapping data of nodes may be expensive in many situations when data contains many fields.

It may be assumed that all keys in the linked list are distinct.

Examples:

```Input : 10->15->12->13->20->14,  x = 12, y = 20
Output: 10->15->20->13->12->14

Input : 10->15->12->13->20->14,  x = 10, y = 20
Output: 20->15->12->13->10->14

Input : 10->15->12->13->20->14,  x = 12, y = 13
Output: 10->15->13->12->20->14```

This may look a simple problem, but is an interesting question as it has the following cases to be handled.

1. x and y may or may not be adjacent.
2. Either x or y may be a head node.
3. Either x or y may be the last node.
4. x and/or y may not be present in the linked list.

How to write a clean working code that handles all the above possibilities.

The idea is to first search x and y in the given linked list. If any of them is not present, then return. While searching for x and y, keep track of current and previous pointers. First change next of previous pointers, then change next of current pointers.

Below is the implementation of the above approach.

## Python

 `# Python program to swap two given nodes ` `# of a linked list` `class` `LinkedList(``object``):` `    ``def` `__init__(``self``):` `        ``self``.head ``=` `None`   `    ``# Head of list` `    ``class` `Node(``object``):` `        ``def` `__init__(``self``, d):` `            ``self``.data ``=` `d` `            ``self``.``next` `=` `None`   `    ``# Function to swap Nodes x and y ` `    ``# in a linked list by changing links` `    ``def` `swapNodes(``self``, x, y):`   `        ``# Nothing to do if x and y are` `        ``# the same` `        ``if` `x ``=``=` `y:` `            ``return`   `        ``# Search for x (keep track of ` `        ``# prevX and CurrX)` `        ``prevX ``=` `None` `        ``currX ``=` `self``.head` `        ``while` `currX !``=` `None` `and` `currX.data !``=` `x:` `            ``prevX ``=` `currX` `            ``currX ``=` `currX.``next`   `        ``# Search for y (keep track of ` `        ``# prevY and currY)` `        ``prevY ``=` `None` `        ``currY ``=` `self``.head` `        ``while` `currY !``=` `None` `and` `currY.data !``=` `y:` `            ``prevY ``=` `currY` `            ``currY ``=` `currY.``next`   `        ``# If either x or y is not present, ` `        ``# nothing to do` `        ``if` `currX ``=``=` `None` `or` `currY ``=``=` `None``:` `            ``return`   `        ``# If x is not head of linked list` `        ``if` `prevX !``=` `None``:` `            ``prevX.``next` `=` `currY`   `        ``else``:  ``# make y the new head` `            ``self``.head ``=` `currY`   `        ``# If y is not head of linked list` `        ``if` `prevY !``=` `None``:` `            ``prevY.``next` `=` `currX` `        ``else``:  `   `            ``# make x the new head` `            ``self``.head ``=` `currX`   `        ``# Swap next pointers` `        ``temp ``=` `currX.``next` `        ``currX.``next` `=` `currY.``next` `        ``currY.``next` `=` `temp`   `    ``# Function to add Node at beginning` `    ``# of list.` `    ``def` `push(``self``, new_data):`   `        ``# 1. alloc the Node and put the data` `        ``new_Node ``=` `self``.Node(new_data)`   `        ``# 2. Make next of new Node as head` `        ``new_Node.``next` `=` `self``.head`   `        ``# 3. Move the head to point to new Node` `        ``self``.head ``=` `new_Node`   `    ``# This function prints contents of ` `    ``# linked list starting from the given Node` `    ``def` `printList(``self``):` `        ``tNode ``=` `self``.head` `        ``while` `tNode !``=` `None``:` `            ``print` `tNode.data,` `            ``tNode ``=` `tNode.``next`   `# Driver code` `llist ``=` `LinkedList()`   `# The constructed linked list is:` `# 1->2->3->4->5->6->7` `llist.push(``7``)` `llist.push(``6``)` `llist.push(``5``)` `llist.push(``4``)` `llist.push(``3``)` `llist.push(``2``)` `llist.push(``1``)` `print` `"Linked list before calling swapNodes() "` `llist.printList()` `llist.swapNodes(``4``, ``3``)` `print` `"` `Linked ``list` `after calling swapNodes() "` `llist.printList()` `# This code is contributed by BHAVYA JAIN`

Output:

```Linked list before calling swapNodes() 1 2 3 4 5 6 7
Linked list after calling swapNodes() 1 2 4 3 5 6 7 ```

Time Complexity: O(n)

Auxiliary Space: O(1)

Optimizations: The above code can be optimized to search x and y in single traversal. Two loops are used to keep program simple.

Simpler approach:

## Python

 `# Python3 program to swap two given` `# nodes of a linked list`   `# A linked list node class` `class` `Node:`   `    ``# constructor` `    ``def` `__init__(``self``, val ``=` `None``, ` `                 ``next1 ``=` `None``):` `        ``self``.data ``=` `val` `        ``self``.``next` `=` `next1`   `    ``# Print list from this` `    ``# to last till None` `    ``def` `printList(``self``):` `        ``node ``=` `self`   `        ``while` `(node !``=` `None``):` `            ``print``(node.data, end ``=` `" "``)` `            ``node ``=` `node.``next` `        ``print``(``" "``)`   `# Function to add a node` `# at the beginning of List` `def` `push(head_ref, new_data):`   `    ``# Allocate node` `    ``(head_ref) ``=` `Node(new_data, head_ref)` `    ``return` `head_ref`   `def` `swapNodes(head_ref, x, y):` `    ``head ``=` `head_ref`   `    ``# Nothing to do if x and y are same` `    ``if` `(x ``=``=` `y):` `        ``return` `None`   `    ``a ``=` `None` `    ``b ``=` `None`   `    ``# Search for x and y in the linked list` `    ``# and store their pointer in a and b` `    ``while` `(head_ref.``next` `!``=` `None``):`   `        ``if` `((head_ref.``next``).data ``=``=` `x):` `            ``a ``=` `head_ref`   `        ``elif` `((head_ref.``next``).data ``=``=` `y):` `            ``b ``=` `head_ref`   `        ``head_ref ``=` `((head_ref).``next``)`   `    ``# If we have found both a and b` `    ``# in the linked list swap current` `    ``# pointer and next pointer of these` `    ``if` `(a !``=` `None` `and` `b !``=` `None``):` `        ``temp ``=` `a.``next` `        ``a.``next` `=` `b.``next` `        ``b.``next` `=` `temp` `        ``temp ``=` `a.``next``.``next` `        ``a.``next``.``next` `=` `b.``next``.``next` `        ``b.``next``.``next` `=` `temp`   `    ``return` `head`   `# Driver code` `start ``=` `None`   `# The constructed linked list is:` `# 1.2.3.4.5.6.7` `start ``=` `push(start, ``7``)` `start ``=` `push(start, ``6``)` `start ``=` `push(start, ``5``)` `start ``=` `push(start, ``4``)` `start ``=` `push(start, ``3``)` `start ``=` `push(start, ``2``)` `start ``=` `push(start, ``1``)`   `print``(``"Linked list before calling swapNodes() "``)` `start.printList()` `start ``=` `swapNodes(start, ``6``, ``1``)` `print``(``"Linked list after calling swapNodes() "``)` `start.printList()` `# This code is contributed by Arnab Kundu`

Output:

```Linked list before calling swapNodes() 1 2 3 4 5 6 7
Linked list after calling swapNodes() 6 2 3 4 5 1 7 ```

Time Complexity: O(n)

Auxiliary Space: O(1)

Please refer complete article on Swap nodes in a linked list without swapping data for more details!

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next