题解 | #链表的奇偶重排#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 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
     */
    ListNode* oddEvenList(ListNode* head) 
    {
        if(head == nullptr)
            return head;
        ListNode* cur = head->next;
        ListNode* tem = head;
        ListNode* curhead = cur;
        while(cur != nullptr && cur->next != nullptr)
        {
            tem->next = cur->next;
            tem = tem->next;

            cur->next = tem->next;
            cur = cur->next;
        }
        tem->next = curhead;
        return head;
    }
};