【Leetcode】：328. Odd Even Linked List 问题 in JAVA

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 ...

题目难度是中等，要求将所有偶数编号的节点放在奇数编号的节点后面，空间复杂度要求为O（1）

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode oddEvenList(ListNode head) {
if (head == null) {
return null;
}
ListNode node = head;
ListNode even = head.next;
while(node.next != null && node.next.next != null) {
ListNode next = node.next;
if (next.next != null) {
node.next = next.next;
next.next = next.next.next;
node = node.next;
} else {
node.next = next.next;
node = node.next;
}
}
if (even != null) {
node.next = even;
}
return head;
}
}