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
## 思路
key: 保存双数结点的首节点 保存单数结点的尾结点
## 代码
```golang
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func oddEvenList(head *ListNode) *ListNode {
if head == nil || head.Next==nil ||head.Next.Next==nil {
return head
}
cur, next := head, head.Next
evenHead, oddTail := next, cur
ct := 1
for next!= nil {
cur.Next = next.Next
cur = next
next = next.Next
ct++
if ct % 2 != 0 {
oddTail = cur
}
}
oddTail.Next = evenHead
return head
}