Merge two sorted linked lists
Write a SortedMerge() function that takes two lists, each of which is sorted in increasing order, and merges the two together into one list which is in increasing order. SortedMerge() should return the new list. The new list should be made by splicing together the nodes of the first two lists.
For example if the first linked list a is 5->10->15 and the other linked list b is 2->3->20, then SortedMerge() should return a pointer to the head node of the merged list 2->3->5->10->15->20.
There are many cases to deal with: either ‘a’ or ‘b’ may be empty, during processing either ‘a’ or ‘b’ may run out first, and finally, there’s the problem of starting the result list empty, and building it up while going through ‘a’ and ‘b’.
Method 1 (Using Dummy Nodes)
The strategy here uses a temporary dummy node as the start of the result list. The pointer Tail always points to the last node in the result list, so appending new nodes is easy.
The dummy node gives the tail something to point to initially when the result list is empty. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’, and adding it to the tail. When
We are done, the result is in dummy.next.
The below image is a dry run of the above approach:
Below is the implementation of the above approach:
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/* C++ program to merge two sorted linked lists */
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class Node
{
public:
int data;
Node* next;
};
/* pull off the front node of
the source and put it in dest */
void MoveNode(Node** destRef, Node** sourceRef);
/* Takes two lists sorted in increasing
order, and splices their nodes together
to make one big sorted list which
is returned. */
Node* SortedMerge(Node* a, Node* b)
{
/* a dummy first node to hang the result on */
Node dummy;
/* tail points to the last result node */
Node* tail = &dummy;
/* so tail->next is the place to
add new nodes to the result. */
dummy.next = NULL;
while (1)
{
if (a == NULL)
{
/* if either list runs out, use the
other list */
tail->next = b;
break;
}
else if (b == NULL)
{
tail->next = a;
break;
}
if (a->data <= b->data)
MoveNode(&(tail->next), &a);
else
MoveNode(&(tail->next), &b);
tail = tail->next;
}
return(dummy.next);
}
/* UTILITY FUNCTIONS */
/* MoveNode() function takes the
node from the front of the source,
and move it to the front of the dest.
It is an error to call this with the
source list empty.
Before calling MoveNode():
source == {1, 2, 3}
dest == {1, 2, 3}
After calling MoveNode():
source == {2, 3}
dest == {1, 1, 2, 3} */
void MoveNode(Node** destRef, Node** sourceRef)
{
/* the front source node */
Node* newNode = *sourceRef;
assert(newNode != NULL);
/* Advance the source pointer */
*sourceRef = newNode->next;
/* Link the old dest off the new node */
newNode->next = *destRef;
/* Move dest to point to the new node */
*destRef = newNode;
}
/* Function to insert a node at
the beginning of the linked list */
void push(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = new 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(Node *node)
{
while (node!=NULL)
{
cout<<node->data<<" ";
node = node->next;
}
}
/* Driver code*/
int main()
{
/* Start with the empty list */
Node* res = NULL;
Node* a = NULL;
Node* b = NULL;
/* Let us create two sorted linked lists
to test the functions
Created lists, a: 5->10->15, b: 2->3->20 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&b, 20);
push(&b, 3);
push(&b, 2);
/* Remove duplicates from linked list */
res = SortedMerge(a, b);
cout << "Merged Linked List is: \n";
printList(res);
return 0;
}
// This code is contributed by rathbhupendra
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/* C program to merge two sorted linked lists */
#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
/* Link list node */
struct Node
{
int data;
struct Node* next;
};
/* pull off the front node of the source and put it in dest */
void MoveNode(struct Node** destRef, struct Node** sourceRef);
/* Takes two lists sorted in increasing order, and splices
their nodes together to make one big sorted list which
is returned. */
struct Node* SortedMerge(struct Node* a, struct Node* b)
{
/* a dummy first node to hang the result on */
struct Node dummy;
/* tail points to the last result node */
struct Node* tail = &dummy;
/* so tail->next is the place to add new nodes
to the result. */
dummy.next = NULL;
while (1)
{
if (a == NULL)
{
/* if either list runs out, use the
other list */
tail->next = b;
break;
}
else if (b == NULL)
{
tail->next = a;
break;
}
if (a->data <= b->data)
MoveNode(&(tail->next), &a);
else
MoveNode(&(tail->next), &b);
tail = tail->next;
}
return(dummy.next);
}
/* UTILITY FUNCTIONS */
/* MoveNode() function takes the node from the front of the
source, and move it to the front of the dest.
It is an error to call this with the source list empty.
Before calling MoveNode():
source == {1, 2, 3}
dest == {1, 2, 3}
After calling MoveNode():
source == {2, 3}
dest == {1, 1, 2, 3} */
void MoveNode(struct Node** destRef, struct Node** sourceRef)
{
/* the front source node */
struct Node* newNode = *sourceRef;
assert(newNode != NULL);
/* Advance the source pointer */
*sourceRef = newNode->next;
/* Link the old dest off the new node */
newNode->next = *destRef;
/* Move dest to point to the new node */
*destRef = newNode;
}
/* Function to insert a node at the beginning of the
linked 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 program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* res = NULL;
struct Node* a = NULL;
struct Node* b = NULL;
/* Let us create two sorted linked lists to test
the functions
Created lists, a: 5->10->15, b: 2->3->20 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&b, 20);
push(&b, 3);
push(&b, 2);
/* Remove duplicates from linked list */
res = SortedMerge(a, b);
printf("Merged Linked List is: \n");
printList(res);
return 0;
}
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/* Java program to merge two
sorted linked lists */
import java.util.*;
/* Link list node */
class Node
{
int data;
Node next;
Node(int d) {data = d;
next = null;}
}
class MergeLists
{
Node head;
/* Method to insert a node at
the end of the linked list */
public void addToTheLast(Node node)
{
if (head == null)
{
head = node;
}
else
{
Node temp = head;
while (temp.next != null)
temp = temp.next;
temp.next = node;
}
}
/* Method to print linked list */
void printList()
{
Node temp = head;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
// Driver Code
public static void main(String args[])
{
/* Let us create two sorted linked
lists to test the methods
Created lists:
llist1: 5->10->15,
llist2: 2->3->20
*/
MergeLists llist1 = new MergeLists();
MergeLists llist2 = new MergeLists();
// Node head1 = new Node(5);
llist1.addToTheLast(new Node(5));
llist1.addToTheLast(new Node(10));
llist1.addToTheLast(new Node(15));
// Node head2 = new Node(2);
llist2.addToTheLast(new Node(2));
llist2.addToTheLast(new Node(3));
llist2.addToTheLast(new Node(20));
llist1.head = new Gfg().sortedMerge(llist1.head,
llist2.head);
llist1.printList();
}
}
class Gfg
{
/* Takes two lists sorted in
increasing order, and splices
their nodes together to make
one big sorted list which is
returned. */
Node sortedMerge(Node headA, Node headB)
{
/* a dummy first node to
hang the result on */
Node dummyNode = new Node(0);
/* tail points to the
last result node */
Node tail = dummyNode;
while(true)
{
/* if either list runs out,
use the other list */
if(headA == null)
{
tail.next = headB;
break;
}
if(headB == null)
{
tail.next = headA;
break;
}
/* Compare the data of the two
lists whichever lists' data is
smaller, append it into tail and
advance the head to the next Node
*/
if(headA.data <= headB.data)
{
tail.next = headA;
headA = headA.next;
}
else
{
tail.next = headB;
headB = headB.next;
}
/* Advance the tail */
tail = tail.next;
}
return dummyNode.next;
}
}
// This code is contributed
// by Shubhaw Kumar
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""" Python program to merge two
sorted linked lists """
# Linked List Node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Create & Handle List operations
class LinkedList:
def __init__(self):
self.head = None
# Method to display the list
def printList(self):
temp = self.head
while temp:
print(temp.data, end=" ")
temp = temp.next
# Method to add element to list
def addToList(self, newData):
newNode = Node(newData)
if self.head is None:
self.head = newNode
return
last = self.head
while last.next:
last = last.next
last.next = newNode
# Function to merge the lists
# Takes two lists which are sorted
# joins them to get a single sorted list
def mergeLists(headA, headB):
# A dummy node to store the result
dummyNode = Node(0)
# Tail stores the last node
tail = dummyNode
while True:
# If any of the list gets completely empty
# directly join all the elements of the other list
if headA is None:
tail.next = headB
break
if headB is None:
tail.next = headA
break
# Compare the data of the lists and whichever is smaller is
# appended to the last of the merged list and the head is changed
if headA.data <= headB.data:
tail.next = headA
headA = headA.next
else:
tail.next = headB
headB = headB.next
# Advance the tail
tail = tail.next
# Returns the head of the merged list
return dummyNode.next
# Create 2 lists
listA = LinkedList()
listB = LinkedList()
# Add elements to the list in sorted order
listA.addToList(5)
listA.addToList(10)
listA.addToList(15)
listB.addToList(2)
listB.addToList(3)
listB.addToList(20)
# Call the merge function
listA.head = mergeLists(listA.head, listB.head)
# Display merged list
print("Merged Linked List is:")
listA.printList()
""" This code is contributed
by Debidutta Rath """
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/* C# program to merge two
sorted linked lists */
using System;
/* Link list node */
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
public class MergeLists
{
Node head;
/* Method to insert a node at
the end of the linked list */
public void addToTheLast(Node node)
{
if (head == null)
{
head = node;
}
else
{
Node temp = head;
while (temp.next != null)
temp = temp.next;
temp.next = node;
}
}
/* Method to print linked list */
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
// Driver Code
public static void Main(String []args)
{
/* Let us create two sorted linked
lists to test the methods
Created lists:
llist1: 5->10->15,
llist2: 2->3->20
*/
MergeLists llist1 = new MergeLists();
MergeLists llist2 = new MergeLists();
// Node head1 = new Node(5);
llist1.addToTheLast(new Node(5));
llist1.addToTheLast(new Node(10));
llist1.addToTheLast(new Node(15));
// Node head2 = new Node(2);
llist2.addToTheLast(new Node(2));
llist2.addToTheLast(new Node(3));
llist2.addToTheLast(new Node(20));
llist1.head = new Gfg().sortedMerge(llist1.head,
llist2.head);
llist1.printList();
}
}
public class Gfg
{
/* Takes two lists sorted in
increasing order, and splices
their nodes together to make
one big sorted list which is
returned. */
public Node sortedMerge(Node headA, Node headB)
{
/* a dummy first node to
hang the result on */
Node dummyNode = new Node(0);
/* tail points to the
last result node */
Node tail = dummyNode;
while(true)
{
/* if either list runs out,
use the other list */
if(headA == null)
{
tail.next = headB;
break;
}
if(headB == null)
{
tail.next = headA;
break;
}
/* Compare the data of the two
lists whichever lists' data is
smaller, append it into tail and
advance the head to the next Node
*/
if(headA.data <= headB.data)
{
tail.next = headA;
headA = headA.next;
}
else
{
tail.next = headB;
headB = headB.next;
}
/* Advance the tail */
tail = tail.next;
}
return dummyNode.next;
}
}
// This code is contributed 29AjayKumar
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<script>
class Node
{
constructor(d)
{
this.data=d;
this.next = null;
}
}
class LinkedList
{
constructor()
{
this.head=null;
}
addToTheLast(node)
{
if (this.head == null)
{
this.head = node;
}
else
{
let temp = this.head;
while (temp.next != null)
temp = temp.next;
temp.next = node;
}
}
printList()
{
let temp = this.head;
while (temp != null)
{
document.write(temp.data + " ");
temp = temp.next;
}
document.write("<br>");
}
}
function sortedMerge(headA,headB)
{
/* a dummy first node to
hang the result on */
let dummyNode = new Node(0);
/* tail points to the
last result node */
let tail = dummyNode;
while(true)
{
/* if either list runs out,
use the other list */
if(headA == null)
{
tail.next = headB;
break;
}
if(headB == null)
{
tail.next = headA;
break;
}
/* Compare the data of the two
lists whichever lists' data is
smaller, append it into tail and
advance the head to the next Node
*/
if(headA.data <= headB.data)
{
tail.next = headA;
headA = headA.next;
}
else
{
tail.next = headB;
headB = headB.next;
}
/* Advance the tail */
tail = tail.next;
}
return dummyNode.next;
}
/* Let us create two sorted linked
lists to test the methods
Created lists:
llist1: 5->10->15,
llist2: 2->3->20
*/
let llist1 = new LinkedList();
let llist2 = new LinkedList();
// Node head1 = new Node(5);
llist1.addToTheLast(new Node(5));
llist1.addToTheLast(new Node(10));
llist1.addToTheLast(new Node(15));
// Node head2 = new Node(2);
llist2.addToTheLast(new Node(2));
llist2.addToTheLast(new Node(3));
llist2.addToTheLast(new Node(20));
llist1.head = sortedMerge(llist1.head,
llist2.head);
document.write("Merged Linked List is:<br>")
llist1.printList();
// This code is contributed by patel2127
</script>
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Output :
Merged Linked List is: 2 3 5 10 15 20Method 2 (Using Local References)
This solution is structurally very similar to the above, but it avoids using a dummy node. Instead, it maintains a struct node** pointer, lastPtrRef, that always points to the last pointer of the result list. This solves the same case that the dummy node did — dealing with the result list when it is empty. If you are trying to build up a list at its tail, either the dummy node or the struct node** “reference” strategy can be used (see Section 1 for details).
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Node* SortedMerge(Node* a, Node* b)
{
Node* result = NULL;
/* point to the last result pointer */
Node** lastPtrRef = &result;
while(1)
{
if (a == NULL)
{
*lastPtrRef = b;
break;
}
else if (b==NULL)
{
*lastPtrRef = a;
break;
}
if(a->data <= b->data)
{
MoveNode(lastPtrRef, &a);
}
else
{
MoveNode(lastPtrRef, &b);
}
/* tricky: advance to point to the next ".next" field */
lastPtrRef = &((*lastPtrRef)->next);
}
return(result);
}
//This code is contributed by rathbhupendra
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|
struct Node* SortedMerge(struct Node* a, struct Node* b)
{
struct Node* result = NULL;
/* point to the last result pointer */
struct Node** lastPtrRef = &result;
while(1)
{
if (a == NULL)
{
*lastPtrRef = b;
break;
}
else if (b==NULL)
{
*lastPtrRef = a;
break;
}
if(a->data <= b->data)
{
MoveNode(lastPtrRef, &a);
}
else
{
MoveNode(lastPtrRef, &b);
}
/* tricky: advance to point to the next ".next" field */
lastPtrRef = &((*lastPtrRef)->next);
}
return(result);
}
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|
Node SortedMerge(Node a, Node b)
{
Node result = null;
/* point to the last result pointer */
Node lastPtrRef = result;
while(1)
{
if (a == null)
{
lastPtrRef = b;
break;
}
else if (b==null)
{
lastPtrRef = a;
break;
}
if(a.data <= b.data)
{
MoveNode(lastPtrRef, a);
}
else
{
MoveNode(lastPtrRef, b);
}
/* tricky: advance to point to the next ".next" field */
lastPtrRef = ((lastPtrRef).next);
}
return(result);
}
// This code contributed by umadevi9616
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|
def SortedMerge( a, b):
result = None;
''' point to the last result pointer '''
lastPtrRef = result;
while(1):
if (a == None):
lastPtrRef = b;
break;
elif(b == None):
lastPtrRef = a;
break;
if(a.data <= b.data):
MoveNode(lastPtrRef, a);
else:
MoveNode(lastPtrRef, b);
''' tricky: advance to point to the next ".next" field '''
lastPtrRef = ((lastPtrRef).next);
return(result);
# This code is contributed by umadevi9616
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|
Node SortedMerge(Node a, Node b)
{
Node result = null;
// Point to the last result pointer
Node lastPtrRef = result;
while(1)
{
if (a == null)
{
lastPtrRef = b;
break;
}
else if (b == null)
{
lastPtrRef = a;
break;
}
if (a.data <= b.data)
{
MoveNode(lastPtrRef, a);
}
else
{
MoveNode(lastPtrRef, b);
}
// tricky: advance to point to
// the next ".next" field
lastPtrRef = ((lastPtrRef).next);
}
return(result);
}
// This code is contributed by gauravrajput1
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|
<script>
function SortedMerge(a,b)
{
let result = null;
/* point to the last result pointer */
let lastPtrRef = result;
while(1)
{
if (a == null)
{
lastPtrRef = b;
break;
}
else if (b==null)
{
lastPtrRef = a;
break;
}
if(a.data <= b.data)
{
MoveNode(lastPtrRef, a);
}
else
{
MoveNode(lastPtrRef, b);
}
/* tricky: advance to point to the next ".next" field */
lastPtrRef = ((lastPtrRef).next);
}
return(result);
}
// This code is contributed by rag2127
</script>
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Method 3 (Using Recursion)
Merge is one of those nice recursive problems where the recursive solution code is much cleaner than the iterative code. You probably wouldn’t want to use the recursive version for production code, however, because it will use stack space which is proportional to the length of the lists.
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Node* SortedMerge(Node* a, Node* b)
{
Node* result = NULL;
/* Base cases */
if (a == NULL)
return(b);
else if (b == NULL)
return(a);
/* Pick either a or b, and recur */
if (a->data <= b->data)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}
// This code is contributed by rathbhupendra
|
|
struct Node* SortedMerge(struct Node* a, struct Node* b)
{
struct Node* result = NULL;
/* Base cases */
if (a == NULL)
return(b);
else if (b==NULL)
return(a);
/* Pick either a or b, and recur */
if (a->data <= b->data)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}
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|
class GFG
{
public Node SortedMerge(Node A, Node B)
{
if(A == null) return B;
if(B == null) return A;
if(A.data < B.data)
{
A.next = SortedMerge(A.next, B);
return A;
}
else
{
B.next = SortedMerge(A, B.next);
return B;
}
}
}
// This code is contributed by Tuhin Das
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|
# Python3 program merge two sorted linked
# in third linked list using recursive.
# Node class
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Constructor to initialize the node object
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
# Method to print linked list
def printList(self):
temp = self.head
while temp :
print(temp.data, end="->")
temp = temp.next
# Function to add of node at the end.
def append(self, new_data):
new_node = Node(new_data)
if self.head is None:
self.head = new_node
return
last = self.head
while last.next:
last = last.next
last.next = new_node
# Function to merge two sorted linked list.
def mergeLists(head1, head2):
# create a temp node NULL
temp = None
# List1 is empty then return List2
if head1 is None:
return head2
# if List2 is empty then return List1
if head2 is None:
return head1
# If List1's data is smaller or
# equal to List2's data
if head1.data <= head2.data:
# assign temp to List1's data
temp = head1
# Again check List1's data is smaller or equal List2's
# data and call mergeLists function.
temp.next = mergeLists(head1.next, head2)
else:
# If List2's data is greater than or equal List1's
# data assign temp to head2
temp = head2
# Again check List2's data is greater or equal List's
# data and call mergeLists function.
temp.next = mergeLists(head1, head2.next)
# return the temp list.
return temp
# Driver Function
if __name__ == '__main__':
# Create linked list :
# 10->20->30->40->50
list1 = LinkedList()
list1.append(10)
list1.append(20)
list1.append(30)
list1.append(40)
list1.append(50)
# Create linked list 2 :
# 5->15->18->35->60
list2 = LinkedList()
list2.append(5)
list2.append(15)
list2.append(18)
list2.append(35)
list2.append(60)
# Create linked list 3
list3 = LinkedList()
# Merging linked list 1 and linked list 2
# in linked list 3
list3.head = mergeLists(list1.head, list2.head)
print(" Merged Linked List is : ", end="")
list3.printList()
# This code is contributed by 'Shriaknt13'.
|
|
using System;
class GFG{
public Node sortedMerge(Node A, Node B)
{
// Base cases
if (A == null)
return B;
if (B == null)
return A;
// Pick either a or b, and recur
if (A.data < B.data)
{
A.next = sortedMerge(A.next, B);
return A;
}
else
{
B.next = sortedMerge(A, B.next);
return B;
}
}
}
// This code is contributed by hunter2000
|
|
<script>
function SortedMerge( A, B) {
if (A == null)
return B;
if (B == null)
return A;
if (A.data < B.data) {
A.next = SortedMerge(A.next, B);
return A;
} else {
B.next = SortedMerge(A, B.next);
return B;
}
}
// This code contributed by umadevi9616
</script>
|
Time Complexity: Since we are traversing through the two lists fully. So, the time complexity is O(m+n) where m and n are the lengths of the two lists to be merged.
Method 4 (Reversing The Lists)
This idea involves first reversing both the given lists and after reversing, traversing both the lists till the end and then comparing the nodes of both the lists and inserting the node with a larger value at the beginning of the result list. And in this way we will get the resulting list in increasing order.
Below is the implementation of above solution.
|
/*Given two sorted linked lists consisting of N and M nodes
respectively. The task is to merge both of the list
(in-place) and return head of the merged list.*/
#include <iostream>
using namespace std;
/* Link list Node */
struct Node {
int key;
struct Node* next;
};
// Function to reverse a given Linked List using Recursion
Node* reverseList(Node* head)
{
if (head->next == NULL)
return head;
Node* rest = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return rest;
}
// Given two non-empty linked lists 'a' and 'b'
Node* sortedMerge(Node* a, Node* b)
{
// Reverse Linked List 'a'
a = reverseList(a);
// Reverse Linked List 'b'
b = reverseList(b);
// Initialize head of resultant list
Node* head = NULL;
Node* temp;
// Traverse both lists while both of them
// have nodes.
while (a != NULL && b != NULL) {
// If a's current value is greater than or equal to
// b's current value.
if (a->key >= b->key) {
// Store next of current Node in first list
temp = a->next;
// Add 'a' at the front of resultant list
a->next = head;
// Make 'a' - head of the result list
head = a;
// Move ahead in first list
a = temp;
}
// If b's value is greater. Below steps are similar
// to above (Only 'a' is replaced with 'b')
else {
temp = b->next;
b->next = head;
head = b;
b = temp;
}
}
// If second list reached end, but first list has
// nodes. Add remaining nodes of first list at the
// beginning of result list
while (a != NULL) {
temp = a->next;
a->next = head;
head = a;
a = temp;
}
// If first list reached end, but second list has
// nodes. Add remaining nodes of second list at the
// beginning of result list
while (b != NULL) {
temp = b->next;
b->next = head;
head = b;
b = temp;
}
// Return the head of the result list
return head;
}
/* Function to print Nodes in a given linked list */
void printList(struct Node* Node)
{
while (Node != NULL) {
cout << Node->key << " ";
Node = Node->next;
}
}
/* Utility function to create a new node with
given key */
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->next = NULL;
return temp;
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* res = NULL;
/* Let us create two sorted linked lists to test
the above functions. Created lists shall be
a: 5->10->15->40
b: 2->3->20 */
Node* a = newNode(5);
a->next = newNode(10);
a->next->next = newNode(15);
a->next->next->next = newNode(40);
Node* b = newNode(2);
b->next = newNode(3);
b->next->next = newNode(20);
cout << "List A before merge: \n";
printList(a);
cout << "\nList B before merge: \n";
printList(b);
/* merge 2 sorted Linked Lists */
res = sortedMerge(a, b);
cout << "\nMerged Linked List is: \n";
printList(res);
return 0;
}
|
|
/*Given two sorted linked lists consisting of N and M nodes
respectively. The task is to merge both of the list
(in-place) and return head of the merged list.*/
import java.util.*;
class GFG{
/* Link list Node */
static class Node {
int key;
Node next;
};
// Function to reverse a given Linked List using Recursion
static Node reverseList(Node head)
{
if (head.next == null)
return head;
Node rest = reverseList(head.next);
head.next.next = head;
head.next = null;
return rest;
}
// Given two non-empty linked lists 'a' and 'b'
static Node sortedMerge(Node a, Node b)
{
// Reverse Linked List 'a'
a = reverseList(a);
// Reverse Linked List 'b'
b = reverseList(b);
// Initialize head of resultant list
Node head = null;
Node temp;
// Traverse both lists while both of them
// have nodes.
while (a != null && b != null) {
// If a's current value is greater than or equal to
// b's current value.
if (a.key >= b.key) {
// Store next of current Node in first list
temp = a.next;
// Add 'a' at the front of resultant list
a.next = head;
// Make 'a' - head of the result list
head = a;
// Move ahead in first list
a = temp;
}
// If b's value is greater. Below steps are similar
// to above (Only 'a' is replaced with 'b')
else {
temp = b.next;
b.next = head;
head = b;
b = temp;
}
}
// If second list reached end, but first list has
// nodes. Add remaining nodes of first list at the
// beginning of result list
while (a != null) {
temp = a.next;
a.next = head;
head = a;
a = temp;
}
// If first list reached end, but second list has
// nodes. Add remaining nodes of second list at the
// beginning of result list
while (b != null) {
temp = b.next;
b.next = head;
head = b;
b = temp;
}
// Return the head of the result list
return head;
}
/* Function to print Nodes in a given linked list */
static void printList(Node Node)
{
while (Node != null) {
System.out.print(Node.key+ " ");
Node = Node.next;
}
}
/* Utility function to create a new node with
given key */
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.next = null;
return temp;
}
/* Driver program to test above functions*/
public static void main(String[] args)
{
/* Start with the empty list */
Node res = null;
/* Let us create two sorted linked lists to test
the above functions. Created lists shall be
a: 5.10.15.40
b: 2.3.20 */
Node a = newNode(5);
a.next = newNode(10);
a.next.next = newNode(15);
a.next.next.next = newNode(40);
Node b = newNode(2);
b.next = newNode(3);
b.next.next = newNode(20);
System.out.print("List A before merge: \n");
printList(a);
System.out.print("\nList B before merge: \n");
printList(b);
/* merge 2 sorted Linked Lists */
res = sortedMerge(a, b);
System.out.print("\nMerged Linked List is: \n");
printList(res);
}
}
// This code is contributed by umadevi9616
|
|
/*Given two sorted linked lists consisting of N and M nodes
respectively. The task is to merge both of the list
(in-place) and return head of the merged list.*/
using System;
using System.Collections.Generic;
public class GFG{
/* Link list Node */
public class Node {
public int key;
public Node next;
};
// Function to reverse a given Linked List using Recursion
static Node reverseList(Node head)
{
if (head.next == null)
return head;
Node rest = reverseList(head.next);
head.next.next = head;
head.next = null;
return rest;
}
// Given two non-empty linked lists 'a' and 'b'
static Node sortedMerge(Node a, Node b)
{
// Reverse Linked List 'a'
a = reverseList(a);
// Reverse Linked List 'b'
b = reverseList(b);
// Initialize head of resultant list
Node head = null;
Node temp;
// Traverse both lists while both of them
// have nodes.
while (a != null && b != null) {
// If a's current value is greater than or equal to
// b's current value.
if (a.key >= b.key) {
// Store next of current Node in first list
temp = a.next;
// Add 'a' at the front of resultant list
a.next = head;
// Make 'a' - head of the result list
head = a;
// Move ahead in first list
a = temp;
}
// If b's value is greater. Below steps are similar
// to above (Only 'a' is replaced with 'b')
else {
temp = b.next;
b.next = head;
head = b;
b = temp;
}
}
// If second list reached end, but first list has
// nodes. Add remaining nodes of first list at the
// beginning of result list
while (a != null) {
temp = a.next;
a.next = head;
head = a;
a = temp;
}
// If first list reached end, but second list has
// nodes. Add remaining nodes of second list at the
// beginning of result list
while (b != null) {
temp = b.next;
b.next = head;
head = b;
b = temp;
}
// Return the head of the result list
return head;
}
/* Function to print Nodes in a given linked list */
static void printList(Node Node)
{
while (Node != null) {
Console.Write(Node.key+ " ");
Node = Node.next;
}
}
/* Utility function to create a new node with
given key */
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.next = null;
return temp;
}
/* Driver program to test above functions*/
public static void Main(String[] args)
{
/* Start with the empty list */
Node res = null;
/* Let us create two sorted linked lists to test
the above functions. Created lists shall be
a: 5.10.15.40
b: 2.3.20 */
Node a = newNode(5);
a.next = newNode(10);
a.next.next = newNode(15);
a.next.next.next = newNode(40);
Node b = newNode(2);
b.next = newNode(3);
b.next.next = newNode(20);
Console.Write("List A before merge: \n");
printList(a);
Console.Write("\nList B before merge: \n");
printList(b);
/* merge 2 sorted Linked Lists */
res = sortedMerge(a, b);
Console.Write("\nMerged Linked List is: \n");
printList(res);
}
}
// This code is contributed by umadevi9616
|
Output:
List A before merge: 5 10 15 40 List B before merge: 2 3 20 Merged Linked List is: 2 3 5 10 15 20 40Time Complexity: Since we are traversing through the two lists fully. So, the time complexity is O(m+n) where m and n are the lengths of the two lists to be merged.
This method is contributed by Mehul Mathur(mathurmehul01).
This idea is similar to this post.
Please refer below post for simpler implementations :
Merge two sorted lists (in-place)
Source: //cslibrary.stanford.edu/105/LinkedListProblems.pdf
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
Merge two sorted linked lists such that merged list is in reverse order
Given two linked lists sorted in increasing order. Merge them such a way that the result list is in decreasing order (reverse order).
Examples:
Input: a: 5->10->15->40 b: 2->3->20 Output: res: 40->20->15->10->5->3->2 Input: a: NULL b: 2->3->20 Output: res: 20->3->21. Using Dummy Nodes
The strategy here uses a temporary dummy node as the start of the result list. The pointer tail always points to the last node in the result list, so appending new nodes is easy. The dummy node gives the tail something to point to initially when the result list is empty. This dummy node is efficient since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either a or b and adding it at the tail. When we are done, the result is in dummy.next.
Following is the C, Java, and Python implementation of the idea:
C
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 | #include <stdio.h> #include <stdlib.h> // A Linked List Node struct Node { int data; struct Node* next; }; // Helper function to print a given linked list void printList(struct Node* head) { struct Node* ptr = head; while (ptr) { printf("%d —> ", ptr->data); ptr = ptr->next; } printf("NULL\n"); } // Helper function to insert a new node at the beginning of the linked list void push(struct Node** head, int data) { struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); newNode->data = data; newNode->next = *head; *head = newNode; } // Function takes the node from the front of the source and moves it // to the front of the destination void moveNode(struct Node** destRef, struct Node** sourceRef) { // if the source list empty, do nothing if (*sourceRef == NULL) { return; } struct Node* newNode = *sourceRef;// the front source node *sourceRef = (*sourceRef)->next;// advance the source pointer newNode->next = *destRef; // link the old dest off the new node *destRef = newNode; // move dest to point to the new node } // Takes two lists sorted in increasing order and merge their nodes // to make one big sorted list, which is returned struct Node* sortedMerge(struct Node* a, struct Node* b) { // a dummy first node to hang the result on struct Node dummy; dummy.next = NULL; // points to the last result node — so `tail->next` is the place // to add new nodes to the result. struct Node* tail = &dummy; while (1) { // if either list runs out, use the other list if (a == NULL) { tail->next = b; break; } else if (b == NULL) { tail->next = a; break; } if (a->data <= b->data) { moveNode(&(tail->next), &a); } else { moveNode(&(tail->next), &b); } tail = tail->next; } return dummy.next; } int main(void) { // input keys int keys[] = { 1, 2, 3, 4, 5, 6, 7 }; int n = sizeof(keys)/sizeof(keys[0]); struct Node *a = NULL, *b = NULL; for (int i = n - 1; i >= 0; i = i - 2) { push(&a, keys[i]); } for (int i = n - 2; i >= 0; i = i - 2) { push(&b, keys[i]); } // print both lists printf("First List: "); printList(a); printf("Second List: "); printList(b); struct Node* head = sortedMerge(a, b); printf("After Merge: "); printList(head); return 0; } |
DownloadRun Code
Output:
First List: 1 —> 3 —> 5 —> 7 —> NULL
Second List: 2 —> 4 —> 6 —> NULL
After Merge: 1 —> 2 —> 3 —> 4 —> 5 —> 6 —> 7 —> NULL
C Exercises: Merge two arrays of same size sorted in decending order
Last update on December 18 2021 11:49:17 (UTC/GMT +8 hours)Python program to merge two lists and sort the merged list
Python program to merge two lists and sort it
In this post, you will learn the Python program to merge two lists and sort them.
A list is a sequence of indexed and ordered values. It is mutable, which means we can change the order of elements in a list. It contains a list of any type of data objects with a comma separated and enclosed within a square bracket. Each element of the list has an index, starting with 0, 1 and so on. For accessing an item, a square bracket is used with the list name.
When we are working with multiple lists, we may come to the situation where we need to merge two lists and sort them. In this post, we will use integer numbers in both lists so that we can easily sort them.