### 题目

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 …

### 直接操作指针

• odd: 指向最后一个奇数位元素。
• even: 指向最后一个偶数位元素。

#### 代码

/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
ListNode odd = new ListNode(0); // dummy odd
ListNode even = new ListNode(0); // dummy even
odd.next = even;
while (cur != null) {
// find next odd
ListNode nextOdd = cur.next;
if (nextOdd != null) { nextOdd = nextOdd.next; }
// insert curr odd
even.next = cur.next;
even = cur.next;
cur.next = odd.next;
odd.next = cur;
odd = cur;
// pass to next odd
cur = nextOdd;
}
odd.next = odd.next.next; // skip dummy even
}
}


### 更优雅的解法

#### 代码

public class Solution {
if (head == null) { return null; }