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OddEvenLinkedList328.java
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OddEvenLinkedList328.java
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/**
* Given a singly linked list, group all odd nodes together followed by the
* even nodes. Please note here we are talking about the node number and not
* the value in the nodes.
*
* You should try to do it in place. The program should run in O(1) space
* complexity and O(nodes) time complexity.
*
* Example:
* Given 1->2->3->4->5->NULL,
* return 1->3->5->2->4->NULL.
*
* Note:
* The relative order inside both the even and odd groups should remain as it was in the input.
* The first node is considered odd, the second node even and so on ...
*
*/
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class OddEvenLinkedList328 {
public ListNode oddEvenList(ListNode head) {
if (head == null) return null;
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode evens = new ListNode(0);
ListNode e = evens;
ListNode p = dummy;
while (p.next != null) {
p = p.next;
if (p.next == null) break;
e.next = new ListNode(p.next.val);
e = e.next;
p.next = p.next.next;
}
p.next = evens.next;
return dummy.next;
}
public ListNode oddEvenList2(ListNode head) {
if (head == null) return null;
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode evens = new ListNode(0);
ListNode e = evens;
ListNode p = dummy;
while (p.next != null) {
p = p.next;
if (p.next == null) break;
e.next = p.next;
e = e.next;
p.next = p.next.next;
e.next = null;
}
p.next = evens.next;
return dummy.next;
}
/**
* https://leetcode.com/problems/odd-even-linked-list/solution/
*/
public ListNode oddEvenList3(ListNode head) {
if (head == null) return null;
ListNode odd = head, even = head.next, evenHead = even;
while (even != null && even.next != null) {
odd.next = even.next;
odd = odd.next;
even.next = odd.next;
even = even.next;
}
odd.next = evenHead;
return head;
}
}