C Program For Swapping Nodes In A Linked List Without Swapping Data - GeeksforGeeks

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. 
x and y may or may not be adjacent.Either x or y may be a head node.Either x or y may be the last node.x and/or y may not be present in the linked list.
x and y may or may not be adjacent.
Either x or y may be a head node.
Either x or y may be the last node.
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 <stdio.h>#include <stdlib.h> // A linked list nodestruct 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 Listvoid 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 listvoid printList(struct Node* node){    while (node != NULL)    {        printf("%d ", node->data);        node = node->next;    }} // Driver codeint 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!
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