[Leetcode 206]链表之单链表反转_golang tmp.next = new(listnode)-优快云博客

本文链接：https://blog.youkuaiyun.com/luis_ora/article/details/82457160

本文提供了使用Python和Golang两种语言实现链表数据结构的方法，并详细展示了如何通过编程技巧完成链表的反转操作。

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Python实现：

用Python实现链表总觉着有点尴尬

class ListNode:
     def __init__(self, x):
         self.val = x
         self.next = None

class Solution:
    def reverseList(self, head):
        if (head is None or head.next is None):
            return head

        head_tmp=None
        tmp=head
        while (tmp !=None):
            next_tmp=tmp.next
            tmp.next=head_tmp
            head_tmp=tmp
            tmp=next_tmp

        return head_tmp

def printlist(head):
    while(head !=None):
        print(head.val)
        head=head.next

p1=ListNode(1)
p2=ListNode(2)
p3=ListNode(3)
p4=ListNode(4)
p1.next=p2
p2.next=p3
p3.next=p4

printlist(p1)
print('**********************')
x=Solution()
pr=x.reverseList(head=p1)
printlist(pr)

Golang实现

package main

import "fmt"

  type ListNode struct {
      Val int
      Next *ListNode
  }

func reverseList(head *ListNode) *ListNode {
   if head==nil || head.Next==nil{
      return head
   }
   var head_tmp *ListNode
   var tmp *ListNode
   var next_tmp *ListNode
   tmp=head
   for tmp !=nil{
      next_tmp=tmp.Next
      tmp.Next=head_tmp
      head_tmp=tmp
      tmp=next_tmp
   }
   return head_tmp

}

func printlist(head *ListNode){
   for {
      if head == nil {
         break

      }

         fmt.Println(head.Val)
         head=head.Next

   }

}

func main() {
   var rl1 *ListNode =new(ListNode)
   var l1 *ListNode = &ListNode{
      Val:1,
      Next:nil,
   }
   var l2 *ListNode = &ListNode{
      Val:2,
      Next:nil,
   }
   var l3 *ListNode = &ListNode{
      Val:3,
      Next:nil,
   }
   var l4 *ListNode = &ListNode{
      Val:4,
      Next:nil,
   }
   l1.Next=l2
   l2.Next=l3
   l3.Next=l4

   printlist(l1)
   fmt.Println("*******")
   rl1=reverseList(l1)
   printlist(rl1)

}