[Leetcode]328. 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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

用两个指针即可，一个指针p指向当前遍历的奇数节点的最后一个位置，另一个指针q指向待提取的奇数节点的前一个位置。

然后把q.next 的节点删除，插入到p.next的位置即可

public class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head != null) {

            ListNode odd = head, even = head.next, evenHead = even; 

            while (even != null && even.next != null) {
                odd.next = odd.next.next; 
                even.next = even.next.next; 
                odd = odd.next;
                even = even.next;
            }
            odd.next = evenHead; 
        }
        return head;
    }

}