Reverse a Linked List | Iterative

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[utkarshsharma]

i tried online articles and famous courses worth 9, 000 to 25, 000. kahi smjh ni aya ki sochna kaise hai, mind mein Linkedlist ke modifications karne kaise hain in c++. but yours' was extraordinary

[manishmotwani]

Nicely explained striver.... You are doing a great job... Keep posting such videos

[krs_rangan]

Please make a series on linked lists, like recursion 🙏

[AnandKumar-kzls]

watched lots of video but this the best one thanks bro

[settyruthvik]

#code in python
class Solution:
# @param A : head node of linked list
# @return the head node in the linked list
def reverseList(self, head):
prev=None
while head:
curr=head
head=head.next
curr.next=prev
prev=curr
return prev

[DeepakGupta-zzvf]

Bro just a request, when u will upload cycle detection question please try to explain why floyd detection works because everyone knows the floyd algorithm but the reasoning is still missing.

[tarunkumar.d]

what's is this Listnode *next line? why are we changing the next value and of which node's next is this? it's just simply Linkedlist *next. is next just like a head pointer???

[pruthvirajjadhav]

Very easy explanation !! I ever watched

[sharonfaithm]

Hands down Striver, YOU are THE BEST.🙌👏

[srinutippana]

6:21
newHead=head
Why is this line needed as we have already pointed head.next=newHead
which will store the reference

[zealkapadiya]

Is it important to know both the recursive and iterative method of this question?

[shaiknoor]

Bro can you solve the below problem
There are n robots with intelligence of k for each robot.
Transfer intelligence so that each and every robot has same intelligence i.e. equal intelligence for each robot .But during transfer X% amount of intelligence get loss.
Input format :-
n X
K1 K2 K3 represents
each robot intelligence respectively i.e k intelligence for each robot
Input:-
3 50
1 2 3
Output:-
1.75, 1.75, 1.75

[jevinmakwana]

Most Optimal Solution:

class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode *newHead = NULL;
while(head!=NULL){
ListNode *next = head->next;
head->next = newHead;
newHead = head;
head=next;
}
return newHead;
}
};

[shlokaggarwal]

We can use recursion. It will be easier.

[kaichang]

understood, thanks for the great explanation

[ranasauravsingh]

UNDERSTOOD...!!!
Thanks, striver for the video... :)

[sanukumar]

why are we creating next node at every iteration, to increase the head pointer we can use head = head.next. but its not working. can anyone explain?

[adi]

Awesome bhaiya.... After linked list what topic should I do bhaiya in dsa c++?

[anonymousaspirant]

why cant we return head instead of dummy(second last step)?