Merging sorted lists

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 21. Merge Two Sorted Lists (EASY)

Linked Lists

#this image is from the net not given in question



You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

 

Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []
Output: []

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

 

Constraints:

  • The number of nodes in both lists is in the range [0, 50].
  • -100 <= Node.val <= 100
  • Both list1 and list2 are sorted in non-decreasing order.

CODE:
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: dummy=curr=ListNode() ''' #curr is a pointer(it is a node acually but we just use
it's next property so it acts like a pointer) of final l_list ,it is initialized by pointing
to dummy node. ''' ''' #dummy node is used to avoid edge case of none type #each element of list is actually a node and has it's value and next assigned
implicitly by leetcode''' while list1 and list2: '''#list 1 and list 2 are heads i.e refer to first node i.e
head=#some node(object of list node class) so actually when we print head it 
prints location of the node object. ''' if list1.val<list2.val: curr.next=list1 list1=list1.next else: curr.next=list2 list2=list2.next curr=curr.next '''# if one list is empty we now add remaining node object to our final l_list ''' if not list1: '''#list 1 is empty i.e it now is equal to None as
list=list1.next(points to None) ''' curr.next=list2 else: curr.next=list1 return dummy.next '''#since we need to return head of our final l_list ,
leetcode will iterate through it and convert it into output form eg[1,1,2,3,4,4] using 
something like this # def display(self): # elems = [ ] # cur_node = self # while cur_node.next != None: # cur_node = cur_node.next # elems.append(cur_node.data) # print(elems). '''

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