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;
}
};

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