LeetCode题解之 Odd Even Linked List

1、题目描述

2、问题分析

将链表拆分成两个，奇数节点形成一个链表，偶数节点形成另外一个链表，最后将偶数节点链表加在奇数节点链表后面。

3、代码

  ListNode* oddEvenList(ListNode* head) {
         if( !head || head->next == NULL  ){
             return head;
         }

         ListNode* odd = head;
         ListNode* even = head->next;
         ListNode* even_m = even ;
         ListNode* p = even->next;
         int count = ;
         while( p != NULL ){
             if( count %  !=  ){
                 odd->next = p ;
                 odd = odd->next;
             }else{
                 even_m->next = p ;
                 even_m = even_m ->next;
             }
             p = p->next;
             ++count ;
         }

         odd->next = even ;
         even_m->next = NULL ;
         return head;

     }