Leetcode 328 Odd Even Linked List 链表

Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

就是将序号为单数的放在前面，而序号为偶数的放在后面

我的方法是讲序号为偶数的的插入到链表末尾。

 /**
  * 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;//空链表返回
         ListNode* last = head;
         int cnt = ;
         for(; last->next; last = last->next, ++cnt);
         if(cnt < ) return head;//链表长度小于3返回
         ListNode* now = head;
         ListNode* end = last;

         for(int i = ; i< cnt/; now = now->next, ++i){

             ListNode* next = now->next;
             now->next = next->next;

             end->next = next;
             next->next = NULL;

             end = next;

         }
         return head;
     }
 };