oddEvenList.go

/* https://leetcode.com/problems/odd-even-linked-list/
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 1:

	Input: 1->2->3->4->5->NULL
	Output: 1->3->5->2->4->NULL

Example 2:

	Input: 2->1->3->5->6->4->7->NULL
	Output: 2->3->6->7->1->5->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 ...
*/

package lll

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func oddEvenList(head *ListNode) *ListNode {
	if head == nil || head.Next == nil {
		return head
	}

	for odd, evenTail := head, head.Next; evenTail != nil && evenTail.Next != nil; odd, evenTail = odd.Next, evenTail.Next {
		cur := evenTail.Next

		evenTail.Next = cur.Next
		cur.Next = odd.Next
		odd.Next = cur
	}
	return head
}