Rearrange a linked list such that all even and odd positioned nodes are together | GeeksforGeeks

[@FrankLi92]

This code can be simplified for readability purposes. There is a lot of variable substitution, that doesn't need to be there.

[@chinmayanand896]

your videos and explaination is too good please make more videos for me...........................

[@piyushmajgawali1611]

if odd is null ,how can you assign odd->next (IN THE FIRST CONDITION TO BREAK)

[@pushp1997]

EXACTLY!

[@jhannyvivek8549]

Well done Harshit Jain...Nicely explained

[@venkateshnaidu2133]

can we do this in O(1) space complexity?

[@bendakai1]

Yes. Keep a pointer at head of the list and move runner pointer 2 nodes forward. detach the node and attach it at head. at eh end we will have all evens at start and odds at other end

[@srinaath9845]

@@bendakai1 nice bro

[@arghadeep10]

luv u harshit muaah ..dhanyabad bhai.

[@GeeksforGeeksVideos]

You're welcome, Arghadeep!

[@spicytuna08]

don't you have to read the data and identify whether or not it is odd or even? here, it seems like the input has to have even number and odd number alternate in order for it to work.

[@cu7695]

He's concerned about position not the exact nature of the data in the node.

[@Shorts-gz3qk]

How come 22 & 70 is odd?

[@spicytuna08]

use C++ list.erase() and list.push_back().

[@reyou7]

cool, good and nice!

[@spicytuna08]

need examples please. that is how most people learn.

[@spicytuna08]

1:12, output is wrong

[@azazahmedtai5730]

he is talknig about positioning/index not about even or odd number

[@aFang-xm9vu]

印度英语听得就是膈应

[@ATULBP1998]

Google translator say "Indian English is the answer"