[数据结构与算法]链表的反转和节点删除

/**

 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverse(struct ListNode* head)
{
    struct ListNode* p=head;
    struct ListNode *temp=NULL;
     struct ListNode *cur2=NULL;
    //struct ListNode *cur1=NULL;
    int i=0;
    if(p==NULL|| p->next==NULL)
        return p;
    while(p){
        temp=p->next;
        p->next=cur2;
        cur2=p;
        p=temp;  
    }
    return cur2;
}

struct ListNode* removeNthFromEnd(struct ListNode* head, int n){
     head= reverse2(head);
    struct ListNode* p=head;
    struct ListNode* temp=NULL;
     struct ListNode* pre=NULL;
    if(head == NULL||n <= 0)
        return head;
        //delete a note
    if(n==1){
        temp=head->next;
        free(head);//链表的删除是否释放内存，head变量还是其内存？
        head=temp;
    }
    while(p && n > 1)
    {
        pre=p;
        p=p->next;
        n--;
    }
    if(pre!=NULL){
        pre->next=p->next;
        free(p);
    }
    head= reverse2( head);   
    return head;
   
}