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 ...

这道题坑得很，他说的是链表元素是第几个这种序号，我愣是没看懂。第一个奇数，第二个偶数，第三个奇数，等等。

不说了代码如下，注意判空就行了。

通常在while循环中判断链表某个值是否不能为空，只需看等式右侧就行了。如下面，等是右侧有对even，even->next 的访问，所以它们绝对不能为空。

/** * 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 == NULL)            return NULL;                ListNode* odd = head;        ListNode* even = head->next;        ListNode* even_head = even;                while(even && even->next != NULL){            odd->next = even->next;            even->next = even->next->next;            even = even->next;            odd = odd->next;        }        odd->next = even_head;                return head;    }};