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题目：
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 ...
思路：
这道题只要将奇数和偶数节点拿出来组成两个链表，然后合并即可。
/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} head * @return {ListNode} */ var oddEvenList = function(head) { if(head==null||head.next==null){ return head; } var p=head.next.next,oHead=head,eHead=head.next,i=3,op=oHead,ep=eHead; while(p!=null){ if(i%2!=0){ op.next=p; op=op.next; }else{ ep.next=p; ep=ep.next; } p=p.next; i++; } ep.next=null; op.next=eHead; return oHead; };
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原文地址：http://www.cnblogs.com/shytong/p/5146752.html
