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 = 1;
        for(; last->next; last = last->next, ++cnt);
        if(cnt < 3) return head;//链表长度小于3返回
        ListNode* now = head;
        ListNode* end = last;
            
        for(int i = 0; i< cnt/2; now = now->next, ++i){
            
            ListNode* next = now->next;
            now->next = next->next;
                
            end->next = next;
            next->next = NULL;
                
            end = next;
            
        }
        return head;
    }
};