[LeetCode] Reverse Linked List I II - 链表翻转问题

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode *first,*second,*third;
    if(head==NULL||head->next==NULL)
        return head;
    first = head;
    second = head->next;
    first->next = NULL;
    while(second!=NULL) {  //注意while(second)不能执行
        third = second->next;
        second->next = first;
        first = second;
        second = third;
    }
    return first;
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
 //个人思路:判断链尾是否存在 翻转到一个新链表
struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode *nhead,*q,*p,*last,*nq,*np;
    int value;
    if(head==NULL||head->next==NULL)
        return head;
    q=head;
    nhead=NULL;     //创建新表头
    while(q->next) {
        //删除最后一个链尾结点
        p=q;
        while(p->next) {
            last=p;
            p=p->next;    
        }
        value=p->val;
        last->next=NULL;
        free(p);
        //插入行结点
        nq=nhead;
        while(nq) {
            last=nq;
            nq=nq->next;
        }
        if(nhead==NULL) { //创建表头
            np=(struct ListNode*)malloc(sizeof(struct ListNode));
            np->val=value;
            nhead=np;
            nhead->next=NULL;
        }
        else {           //插入结点
            np=(struct ListNode*)malloc(sizeof(struct ListNode));
            np->val=value;
            np->next=NULL;
            last->next=np;
        }
        //q结点循环前始终指向链表头
        q=head;
    }
    //最后一个结点及头结点head
    nq=nhead;
    while(nq) {
        last=nq;       //使用nq=np总是报错WR
        nq=nq->next;
    }
    np=(struct ListNode*)malloc(sizeof(struct ListNode));
    np->val=head->val;
    np->next=NULL;
    last->next=np;    //nq->next=np会报错RE 因为nq此时为next及null,而nq->next更不知道在哪
    return nhead;
}