给你单链表的头节点 head ,请你反转链表,并返回反转后的链表。
示例 1:
输入:head = [1,2,3,4,5]
输出:[5,4,3,2,1]
示例 2:
输入:head = [1,2]
输出:[2,1]
示例 3:
输入:head = []
输出:[]
分析:
- 思路一: 设置三个指针
代码实现:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* reverseList(struct ListNode* head){
//当为空链表的时候:
if(head==NULL)
return NULL;
struct ListNode *n1=NULL;
struct ListNode *n2=head;
struct ListNode *n3=head->next;
while(n2){
//反转
n2->next=n1;
//迭代
n1=n2;
n2=n3;
if(n3)
n3=n3->next;
}
return n1;
}
**思路二:**头插法
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* reverseList(struct ListNode* head){
//当为空链表的时候:
if(head==NULL)
return NULL;
struct ListNode *newnode=NULL;
struct ListNode *n1=head;
struct ListNode *n2=head->next;
while(n1){
//反转
n1->next=newnode;
newnode=n1;
//迭代
n1=n2;
if(n2)
n2=n2->next;
}
return newnode;
}
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* reverseList(struct ListNode* head){
//当为空链表的时候:
if(head==NULL)
return NULL;
struct ListNode *newnode=NULL;
struct ListNode *cur=head;
while(cur){
struct ListNode *next=cur->next;
//反转
cur->next=newnode;
newnode=cur;
//迭代
cur=next;
}
return newnode;
}