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; } 
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