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 = 1;
        while( p != NULL ){
            if( count % 2 != 0 ){
                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;
        
        
    }