Merge sort Linked list

#1

what I am missing here?

#2

Hello @Vikaspal,

Mistakes:

  1. Line 81 and line 88
  2. Line 97
  3. Line 109

Corrected Code:

Hope, this would help.
Give a like if you are satisfied.

1 Like
#3

@S18ML0016

  1. line 97 . I can’t check like this if (head == NULL || head->next == NULL ) when i have the single node and also no node
#4

Hello @Vikaspal,

Your problem is not clear. Please, give an example in support of your doubt.

#5

@S18ML0016 you mentions that i missed the case when there is only single node i combine that both the case when there is no node head=null or when there is single node head->next == null . I want to put the conditions explicitly?

#6

Hello @Vikaspal,

You can skip the case of head==NULL here, correcting myself.

Reason:
you will never reach this base case as you are returning for single node i.e. head->next==NULL.

Hope, it will help.

#7

@S18ML0016 so it works fine for the both the even head is null first it will check it if i don’t write head == null it may happen head->next means null->next may create some issue ?

#8

Hey @Vikaspal,

That is the catch, this will never happen.

Reason:
for array of size 3:
Division: one of size 1(return) and another of size 2.

for array of size 2:
Division: one of size 1(return) and another of size 1(return).

for array of size 1:
return

So, you’ll never reach the case of head == NULL.

Hope, it is clear now.

#9

@S18ML0016 got it ur points

#10

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