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 …

原题链接：https://leetcode.com/problems/odd-even-linked-list/

思路
奇节点的下一个节点恰好就是偶节点的下一个节点（1的下一个节点应该是2的下一个节点，也就是3）;
而偶节点的下一个节点也恰好是奇节点是下一个节点
所以只需要odd even两个循环往下遍历就可以了，最后把奇链表的尾部指向偶链表的头。

代码（C++）

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* oddEvenList(ListNode* head)     {        if (!head)             return head;        ListNode* odd = head;        ListNode* even = head->next;        ListNode* evenHead = even;        while (odd->next && even->next)         {            odd->next = even->next;            odd = odd->next;            even->next = odd->next;            even = even->next;        }        odd->next = evenHead;        return head;    }};