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

• Last Updated : 30 Mar, 2022

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.

## C

 `// C program to swap the nodes of linked list``// rather than swapping the field from the nodes.``#include ``#include ` `// A linked list node``struct` `Node``{``    ``int` `data;``    ``struct` `Node* next;``};` `/* Function to swap nodes x and y in linked list``   ``by changing links */``void` `swapNodes(``struct` `Node** head_ref,``               ``int` `x, ``int` `y)``{``    ``// Nothing to do if x and y are same``    ``if` `(x == y)``        ``return``;` `    ``// Search for x (keep track of prevX and CurrX``    ``struct` `Node *prevX = NULL, *currX = *head_ref;``    ``while` `(currX && currX->data != x)``    ``{``        ``prevX = currX;``        ``currX = currX->next;``    ``}` `    ``// Search for y (keep track of prevY and CurrY``    ``struct` `Node *prevY = NULL, *currY = *head_ref;``    ``while` `(currY && currY->data != y)``    ``{``        ``prevY = currY;``        ``currY = currY->next;``    ``}` `    ``// If either x or y is not present,``    ``// nothing to do``    ``if` `(currX == NULL || currY == NULL)``        ``return``;` `    ``// If x is not head of linked list``    ``if` `(prevX != NULL)``        ``prevX->next = currY;``    ``else` `        ``// Else make y as new head``        ``*head_ref = currY;` `    ``// If y is not head of linked list``    ``if` `(prevY != NULL)``        ``prevY->next = currX;``    ``else``       ` `        ``// Else make x as new head``        ``*head_ref = currX;` `    ``// Swap next pointers``    ``struct` `Node* temp = currY->next;``    ``currY->next = currX->next;``    ``currX->next = temp;``}` `// Function to add a node at the``// beginning of List``void` `push(``struct` `Node** head_ref,``          ``int` `new_data)``{``    ``// Allocate node``    ``struct` `Node* new_node =``          ``(``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));` `    ``// Put in the data``    ``new_node->data = new_data;` `    ``// Link the old list off the new node``    ``new_node->next = (*head_ref);` `    ``// Move the head to point to the new node``    ``(*head_ref) = new_node;``}` `// Function to print nodes in a given``// linked list``void` `printList(``struct` `Node* node)``{``    ``while` `(node != NULL)``    ``{``        ``printf``(``"%d "``, node->data);``        ``node = node->next;``    ``}``}` `// Driver code``int` `main()``{``    ``struct` `Node* start = NULL;` `    ``// The constructed linked list is:``    ``// 1->2->3->4->5->6->7``    ``push(&start, 7);``    ``push(&start, 6);``    ``push(&start, 5);``    ``push(&start, 4);``    ``push(&start, 3);``    ``push(&start, 2);``    ``push(&start, 1);` `    ``printf``(``"Linked list before calling swapNodes() "``);``    ``printList(start);` `    ``swapNodes(&start, 4, 3);` `    ``printf``(``"Linked list after calling swapNodes() "``);``    ``printList(start);` `    ``return` `0;``}`

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)

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

My Personal Notes arrow_drop_up