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 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 interesting question as it has 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 last node.
4) x and/or y may not be present in linked list.
How to write a clean working code that handles all of the above possibilities.
The idea it to first search x and y in 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. Following are C and Java implementations of this approach.
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C Programming:
/* This program swaps the nodes of linked list rather
than swapping the field from the nodes.*/
#include<stdio.h>
#include<stdlib.h>
/* 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 begining 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;
}
}
/* Druver program to test above function */
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("\n Linked list before calling swapNodes() ");
printList(start);
swapNodes(&start, 4, 3);
printf("\n 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
Optimizations: The above code can be optimized to search x and y in single traversal. Two loops are used to keep program simple.
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