# Swap 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 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

We strongly recommend to minimize your browser and try this yourself first.

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.

## C

```/* 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
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 (prevX != NULL)
prevX->next = currY;
else // Else make y as new head

if (prevY != NULL)
prevY->next = currX;
else  // Else make x as new head

// 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 */

/* move the head to point to the 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;
}
```

## Java

```// Java program to swap two given nodes of a linked list

class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}

{

/* Function to swap Nodes x and y in linked list by
public void swapNodes(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)
Node prevX = null, currX = head;
while (currX != null && currX.data != x)
{
prevX = currX;
currX = currX.next;
}

// Search for y (keep track of prevY and currY)
Node prevY = null, currY = head;
while (currY != null && 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 (prevX != null)
prevX.next = currY;
else //make y the new head

if (prevY != null)
prevY.next = currX;
else // make x the new head

// Swap next pointers
Node temp = currX.next;
currX.next = currY.next;
currY.next = temp;
}

/* Function to add Node at beginning of list. */
public void push(int new_data)
{
/* 1. alloc the Node and put the data */
Node new_Node = new Node(new_data);

/* 2. Make next of new Node as head */

/* 3. Move the head to point to new Node */
}

/* This function prints contents of linked list starting
from the given Node */
public void printList()
{
while (tNode != null)
{
System.out.print(tNode.data+" ");
tNode = tNode.next;
}
}

/* Druver program to test above function */
public static void main(String[] args)
{

/* 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);

System.out.print("\n Linked list before calling swapNodes() ");
llist.printList();

llist.swapNodes(4, 3);

System.out.print("\n Linked list after calling swapNodes() ");
llist.printList();
}
}
// This code is contributed by Rajat Mishra
```

## Python

```
# Python program to swap two given nodes of a linked list
def __init__(self):

class Node(object):
def __init__(self, d):
self.data = d
self.next = None

# Function to swap Nodes x and y in linked list by
def swapNodes(self, x, y):

# Nothing to do if x and y are same
if x == y:
return

# Search for x (keep track of prevX and CurrX)
prevX = None
while currX != None and currX.data != x:
prevX = currX
currX = currX.next

# Search for y (keep track of prevY and currY)
prevY = None
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 prevX != None:
prevX.next = currY
else: #make y the new head

if prevY != None:
prevY.next = currX
else: # make x the new head

# 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

# 3. Move the head to point to new Node

# This function prints contents of linked list starting
# from the given Node
def printList(self):
while tNode != None:
print tNode.data,
tNode = tNode.next

# Driver program to test above function

# 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 "\nLinked 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```

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 –

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

```#include <iostream>

using namespace std;

// A linked list node class
class Node {

public:
int data;
class Node* next;

// constructor
Node(int val, Node* next)
: data(val), next(next)
{
}

// print list from this
// to last till null
void printList()
{

Node* node = this;

while (node != NULL) {

cout << node->data;
node = node->next;
}

cout << endl;
}
};

// Function to add a node
// at the beginning of List
{

// allocate node
}

void swap(Node*& a, Node*& b)
{

Node* temp = a;
a = b;
b = temp;
}

void swapNodes(Node** head_ref, int x, int y)
{

// Nothing to do if x and y are same
if (x == y)
return;

Node **a = NULL, **b = NULL;

// search for x and y in the linked list
// and store therir pointer in a and b

}

else if ((*head_ref)->data == y) {
}

}

// if we have found both a and b
// in the linked list swap current
// pointer and next pointer of these
if (a && b) {

swap(*a, *b);
swap(((*a)->next), ((*b)->next));
}
}

int main()
{

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);

cout << "Linked list before calling swapNodes() ";
start->printList();

swapNodes(&start, 6, 3);

cout << "Linked list after calling swapNodes() ";
start->printList();
}
```

Output:

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

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