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
}