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.

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 ...

给定一个桉顺序排列的数组（从1开始），把该数组奇数在前，偶数在后排列（奇数和偶数的顺序需要和原链表中数字出现顺序相同）

分成两个链表，第一个存奇数，第二个存偶数

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode OddEvenList(ListNode head) {
        if(head == null)
        {
            return head;
        }
        ListNode evenHead = head.next;
        ListNode oddCur = head;
        ListNode evenCur = head.next;
        
        while(evenCur != null && evenCur.next != null)
        {
            oddCur.next = evenCur.next;
            oddCur = oddCur.next;
            
            evenCur.next = oddCur.next;
            evenCur = evenCur.next;
        }
        
        oddCur.next = evenHead;
        return head;
    }
}