题解 | #链表的奇偶重排#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @return ListNode类
 */
struct ListNode* oddEvenList(struct ListNode* head)
{
	if (head == NULL || head->next == NULL || head->next->next == NULL)
		return head;
	struct ListNode* odd = head;
	struct ListNode* neweven = head->next;
	struct ListNode* even = head->next;
	struct ListNode* nextodd = odd->next->next;
	struct ListNode* nexteven = even->next->next;
	while (even != NULL && even->next != NULL)
	{
		odd->next = nextodd;
		even->next = nexteven;
		odd = nextodd;
		even = nexteven;
		if (nexteven != NULL && nexteven->next != NULL)
		{
			nextodd = nextodd->next->next;
			nexteven = nexteven->next->next;
		}
	}
	odd->next=neweven;
	if(even)
		even->next=NULL;
	return head;
}