leetcode 206. Reverse Linked List 翻转链表
文章目录
-
- 迭代：
-
- C++
- Java
- Python
- Go
- 递归：
-
- C++
- Java
- Python
- Go
Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?
迭代：
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* pre=NULL;
ListNode* cur=head;
while(cur!=NULL){
ListNode* tmp=cur->next;
cur->next=pre;
pre=cur;
cur=tmp;
}
return pre;
}
};
- 还可以这样写：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head==NULL||head->next==NULL)return head;
ListNode* dummy=new ListNode(-1);
dummy->next=head;
ListNode* pre=dummy,*cur=head,*nxt;
while(cur->next!=NULL){
nxt=cur->next;
cur->next=nxt->next;
nxt->next=pre->next;
pre->next=nxt;
}
return dummy->next;
}
};
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode pre=null,cur=head,tmp;
while(cur!=null){
tmp=cur.next;
cur.next=pre;
pre=cur;
cur=tmp;
}
return pre;
}
}
Python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
pre,cur=None,head
while cur is not None:
nxt=cur.next
cur.next=pre
pre=cur
cur=nxt
return pre
- 简化写法：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
pre,cur=None,head
while cur:
cur.next,pre,cur=pre,cur,cur.next
return pre
Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
var pre *ListNode
cur:=head
for cur!=nil {
cur.Next,pre,cur=pre,cur,cur.Next
}
return pre
}
递归：
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head==NULL||head->next==NULL)return head;
ListNode* p=reverseList(head->next);
head->next->next=head;
head->next=NULL;
return p;
}
};
Java
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if(head==null||head.next==null)return head;
ListNode p=reverseList(head.next);
head.next.next=head;
head.next=null;
return p;
}
}
Python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
p=self.reverseList(head.next)
head.next.next=head
head.next=None
return p
Go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
if head==nil || head.Next==nil {
return head
}
p:=reverseList(head.Next)
head.Next.Next=head
head.Next=nil
return p
}