力扣刷题—链表反转

法一：反转指针

 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* reverseList(struct ListNode* head){
if(head==NULL)
return NULL;

struct ListNode*n1=NULL;//n1作为返回的头节点
struct ListNode*n2=head;
struct ListNode*n3=n2->next;//记录头节点的下一个节点，避免转过去之后找不到后面的节点


    while(n2)//n2是反转的主角，所以n2到空才可以，迭代过程
    {
        n2->next=n1;
        n1=n2;
        n2=n3;
        if(n3)
        n3=n3->next;
    }
    return n1;


}

 法二：头插法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* reverseList(struct ListNode* head){
if(head==NULL)
return NULL;
struct ListNode*cur=head;
struct ListNode*newnode=NULL;
struct ListNode*next=cur->next;
while(cur)
{
    cur->next=newnode;
    newnode=cur;
    cur=next;
    if(next)
    next=next->next;

}
return newnode;

}