leetcode 328. Odd Even Linked List

题目内容

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 …

题目分析

给定一个单链表，重新链接，将所有奇数位的结点相连，在后边再接偶数位的所有结点。

所有的链表都是从奇数，开始的，由例子可以看出。

本题的一个比较重要的地方是先保留偶数结点的第一个结点的指向。

然后就是不断的next指下去。

/**
* 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) return head;

ListNode odd=head;
ListNode even=head.next;
ListNode evenhead=head.next;

while(even!=null&&even.next!=null)
{
odd.next=even.next;
odd=odd.next;
even.next=odd.next;
even=even.next;
}
odd.next=evenhead;

return head;
}

}