LeetCode题解：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 …

题意：将单链表结构变为前半部分是奇数，后半部分是偶数，处理后数据的顺序不变

解题思路：双指针，一个用于奇数，一个用于偶数，最终合并就好了。

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) {            ListNode odd = head, even = head.next, evenHead = even;             while (even != null && even.next != null) {                odd.next = odd.next.next;                 even.next = even.next.next;                 odd = odd.next;                even = even.next;            }            odd.next = evenHead;         }        return head;    }}