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

.

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 ...

分析：本题的注意点在题目中已经提到：根据结点位置的奇偶来重排，而不是结点的数据。

思路：将奇结点和偶结点分开形成两的链表，然后把两个链表头尾相连。

代码：

/**
* 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 || head.next == null || head.next.next == null) return head;//直接返回的三种情况
ListNode odd = head;// 指向奇节点
ListNode even = head.next;// 指向偶结点的头
ListNode evenHead = head.next;// 指向偶结点的头
while(odd.next != null && even.next != null) {
odd.next = even.next; // 奇指针next指向偶指针的下一个元素（即<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">奇指针的next</span>指向下一个奇结点）
odd = odd.next;       // 奇指针后移（指向下一个奇结点）
even.next = odd.next; // <span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">偶指针next指向奇指针的下一个元素（即</span><span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">偶指针的next</span><span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">指向下一个偶结点）</span>
even = even.next;     // 偶指针后移（指向下一个偶结点）
}// 跳出循环时，odd 和 even 分别指向奇结点链表和偶结点链表的末尾
odd.next = evenHead;// 奇结点链表的末尾与偶结点链表的头相连
return head;
}
}