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 …

s思路:
1. 这个简单啊。由于是分成两个链表后再连接起来，那么先找到两个链表的头，然后按照交叉的顺序一步一步的分解成两条链表，由于是很规则的操作，所以也没啥幺蛾子。

class Solution {public:    ListNode* oddEvenList(ListNode* head) {        //        if(!head) return NULL;        ListNode* head1=head,*head2=head->next;//bug:看到head->next，就要自然反映这是在裸奔，没有加保护的裸奔，所以要把保护加在前面！        ListNode* odd=head1,*even=head2;        while(even&&even->next){//注意：            odd->next=even->next;                odd=odd->next;            if(odd){                even->next=odd->next;                even=even->next;                }        }        odd->next=head2;        return head;    }};