merge two sorted lists
merge two sorted lists - leetcode
merge two sorted lists python
merge two sorted doubly linked lists
merge two sorted linked list without duplicates
merge two sorted linked lists python

I wrote such a solution to `merge two sorted list`

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

```Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
```

My solution:

```# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution3:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
"""
Plan:
Compare l1 and l2 and merge the remainders
"""
l3 = ListNode(0)

while l1 and l2: #assert both exist
if l2.val < l1.val:
l3 = l2 #build l3's node
l2 = l2.next #this is i++
else:
l3 = l1
l1 = l1.next
l3 = l3.next #find the next to build
if l1:
l3 = l1
if l2:
l3 = l2

```

Input

[1,2,4] [1,3,4]

Output

[0]

Expected

[1,1,2,3,4,4]

I checked but cannot find any problems with my logic.

Your `l3` is supposed to identify the place where a next node shall be appended, hence

```   l3 = head
```

You need to append the head of one of lists given (`l1` or `l2`) to `l3`, but you don't. And after you append the node you need to advance both the head pointer of the source list (`l1` or `l2`) and the tail pointer of the destination list (`l3`):

```    while l1 and l2:     #assert both exist
if l2.val < l1.val:
l3.next = l2 #append new node to the resulting list
l2 = l2.next #and skip it in the source list
else:
l3.next = l1
l1 = l1.next
l3 = l3.next     #find the next to build
```

Merging two sorted linked lists, That is, your returned list will never contain more than two nodes, regardless of input It's been a while So far all of the answers have been interesting and well done. It's possible that 2 Down vote. So merging polygen wit AndreyT we get: Problem statement. Given 2 sorted singly linked list. Write a function to merge given two sorted linked lists. List1: 10->15->17->20 List2: 5->9->13->19 Result: 5->9->10->13->15->17->19->20

here is my solution

```class ListNode:
def __init__(self, x):
self.val = x
self.next = None

def insert(self,el):
Node = self
while Node:
if Node.next==None:
Node.next = ListNode(el)
break
else:
Node =Node.next

def prin(node):
while node:
if node.next:
print(node.val,end='-> ')
else:
print(node.val)
node = node.next

def mergeTwoLists(l1: ListNode, l2: ListNode) -> ListNode:
"""
Plan:
Compare l1 and l2 and merge the remainders
"""
_node = None
if l1.val<l2.val:
_node = ListNode(l1.val)
l1 =l1.next
else:
_node = ListNode(l2.val)
l2=l2.next

l3 = _node

while l1 and l2: #assert both exist
if l2.val < l1.val:
l3.insert(l2.val)
l2 = l2.next
else:
l3.insert(l1.val)
l1 = l1.next

while l1:
l3.insert(l1.val)
l1 =l1.next

while l2:
l3.insert(l2.val)
l2 = l2.next

return l3

node1= ListNode(1)
node1.insert(2)
node1.insert(4)

node2 = ListNode(1)
node2.insert(3)
node2.insert(4)

solved_list = mergeTwoLists(node1,node2)
solved_list.prin()
```

Merge two sorted linked lists, Write a SortedMerge() function that takes two lists, each of which is sorted in For example if the first linked list a is 5->10->15 and the other linked list b is Get hold of all the important DSA concepts with the DSA Self Paced� You’re given the pointer to the head nodes of two sorted linked lists. The data in both lists will be sorted in ascending order. Change the next pointers to obtain a single, merged linked list which also has data in ascending order. Either head pointer given may be null meaning that the corresponding list is empty.

Your function returns `head.next`, which is only modified in the following lines:

```l3 = ListNode(0) # this is what's returned
... # rest of the code changes l3, but never touches head.next
```

You seem to assume that, with this setup, any assignments to `l3` will modify `head.next`, which is not correct. As it stands, the function will always return the newly created node with content 0.

A more correct approach would be:

```# at the beginning
...
l3.next = newnode
l3 = l3.next # moving the pointer to the newly added tail
```

Amazon Coding Interview Question, Merge Two Sorted Lists is a coding interview question asked by Amazon (and In this video Duration: 14:58 Posted: Feb 15, 2020 I have a function that correctly returns the multiples of 3 and those of 5 in 2 separate linked lists. Now I want to combine those 2 Linked Lists into a single sorted non-repeating Linked List. I tried the following code, but "next" is not a defined element in my case since I'm using the Linked List library.

You're never changing the list structure (there are no assignments to any `next`s in the loop), you're just moving `l1`, `l2`, and `l3` along the input lists. Then you return `head.next`, which is still the node you first assigned to it.

Since your approach with a sentinel node makes the code much simpler than without, I've kept it here:

```def merge(l1, l2):
first = ListNode(0)
# 'here' is where we link in the nodes.
here = first
while l1 and l2:
# Pick the next node from the inputs.
# Note that this has the same effect as assigning
# to 'first.next' on the first iteration, when 'first' and 'here'
# refer to the same object.
if l1.val < l2.val:
here.next = l1
else:
here.next = l2
# Move along to the last node we picked.
here = here.next
# Grab the rest.
if l1:
here.next = l1
else:
here.next = l2
return first.next
```

Using Merge Sort to Sort a Linked List, Sort the sublists recursively and then merge the sorted lists together to form the answer. STEP 2: Divide the linked list into two equal halves. sort the sublists MergeSort(&b); *headRef = SortedMerge(a, b); // answer = merge the two sorted lists together } SURPRISE ME WITH More Awesome Stuff. Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4

Quick solution:

```class ListNode:
def __init__(self, x):
self.val = x
self.next = None

class Solution3:
def merge_two_lists(self, l1: ListNode, l2: ListNode):

l3 = ListNode(None)

while l1 and l2:
if l2.val < l1.val:
l3.next = l2
l2 = l2.next
else:
l3.next = l1
l1 = l1.next
l3 = l3.next
if l1:
l3.next = l1
if l2:
l3.next = l2

while no:
no = no.next
```

Output:

```Singly-linked list: 1
```

What is the shortest function that merges two sorted lists of length 2 , Is using insertion sort on a linked list faster than using it on an array? Originally Answered: What is the shortest algorithm to define a function that merges two is even simpler and shorter, but sorted() has already been banned for some� Recursive algorithm for merging two sorted linked lists. def merge_lists(h1, h2): if h1 is None: return h2 if h2 is None: return h1 if (h1.value < h2.value): h1.next = merge_lists(h1.next, h2) return h1 else: h2.next = merge_lists(h2.next, h1) return h2. share.

• In your loop you never set the `.next` so what did you expect to obtain? You simply give different names to different objects without ever doing anything to them.