Linked List Cycle-II #142

Reaper — Thu, 03 Jul 2025 17:53:38 +0000

So basically use slow and fast pointers to detect cycle if slow pointer and fast pointer meet each other it means there's a cycle.

and after that to determine the starting of the cycle, just start a pointer from the head and move it until its not equal to the slow pointer, ps: move both slow and start pointer altogether until they meet

when they meet return either of them!

Odd-Even Linked List solution #leetcode328

Reaper — Thu, 03 Jul 2025 15:19:22 +0000

Intuition: Make two dummy nodes OddHead and EvenHead, all the odd indices goes to OddHead, and even indices go to EvenHead.

Approach:
OddHead will have all the odd ones, EvenHead will have all the even ones
connect them via a pointer.

Java Solution:-

public ListNode oddEvenList(ListNode head){
    if(head == null || head.next == null) return head;
    ListNode oddHead = new ListNode(-1);
    ListNode evenHead = new ListNode(-1);
    ListNode oddptr = oddHead;
    ListNode evenptr = evenHead;
    ListNode current = head;
    int index = 1;
    while(current!= null){
       if(index%2 == 1){
         oddptr.next = current;
         oddptr = oddptr.next;
        }
        else{
          evenptr.next = current;
          evenptr = evenptr.next;
        }
        current = current.next;
        index++;
     }
     evenptr.next = null;
     oddptr.next = evenHead.next;
     return oddHead.next;
}

DEV Community: Reaper

Linked List Cycle-II #142

Odd-Even Linked List solution #leetcode328