leetcode笔记—Reorder List

给一个链表 L: L0→L1→…→Ln-1→Ln,

需要返回: L0→Ln→L1→Ln-1→L2→Ln-2→…

思路1：

void reorderList(ListNode* head) {
       ListNode* p=head;
        while(p&&p->next)
        {
            ListNode* pnext=p->next;   //记下下一个节点
            ListNode* temp=p;
            if(p->next->next==NULL)
            {
                p=p->next;
            }
            else
            {
                while(temp->next&&temp->next->next)
                {
                    temp=temp->next;

                }//让temp指向倒数第二个节点
                p->next=temp->next;  //指向最后一个节点
                temp->next=NULL;    //删除最后一个节点
                p=p->next;          //连接后面的节点
                p->next=pnext;
                p=p->next;

            }
        }
        if(p)
        {
            p->next=NULL;
     }}

思路2：

if(head == NULL){
            return;
        }  

        ListNode* p1 = head;
        ListNode* p2 = splitList(head);  

        p2 = revertList(p2);  

        mergeList(p1, p2);
    }  

    void mergeList(ListNode * p1, ListNode * p2){  //合并链表
        while(p2 != NULL){
            ListNode* tmp = p2;
            p2 = p2->next;  

            tmp->next = p1->next;
            p1->next = tmp;
            p1 = p1->next;
            p1 = p1->next;
        };
    }  

    ListNode* splitList(ListNode *head){  //去后半部分链表
        ListNode* slow = new ListNode(0);
        slow->next = head;
        ListNode* fast = slow;  

        while(fast->next != NULL && fast->next->next != NULL){
            slow = slow->next;
            fast = fast->next;
            fast = fast->next;
        }  

        if(fast->next != NULL){
            slow = slow->next;
            fast = fast->next;
        }  

        ListNode* tmp = slow->next;
        slow->next = NULL;
        return tmp;
    }  

    ListNode* revertList(ListNode* head){  //翻转链表
        if(head == NULL){
            return NULL;
        }  

        ListNode* p = head->next;
        head->next = NULL;  

        while(p != NULL){
            ListNode* tmp = p;
            p = p->next;
            tmp->next = head;
            head = tmp;
        }
        return head;
    }