在Go（Golang）中逆转一个链表的方法

概述

其目的是逆转一个给定的链表。

实验结果

Input:  3->2->1
Output: 1->2->3

程序

下面是一个同样的程序

package main

import "fmt"

func main() {
	first := initList()
	first.AddFront(1)
	first.AddFront(2)
	first.AddFront(3)
	first.AddFront(4)

	first.Head.Traverse()
	first.Reverse()
        fmt.Println("")
	first.Head.Traverse()

}

func initList() *SingleList {
	return &SingleList{}
}

type ListNode struct {
	Val  int
	Next *ListNode
}

func (l *ListNode) Traverse() {
	for l != nil {
		fmt.Println(l.Val)
		l = l.Next
	}
}

type SingleList struct {
	Len  int
	Head *ListNode
}

func (s *SingleList) Reverse() {

	curr := s.Head
	var prev *ListNode
	var next *ListNode

	for curr != nil {
		next = curr.Next
		curr.Next = prev
		prev = curr
		curr = next
	}
	s.Head = prev
}
func (s *SingleList) AddFront(num int) {
	ele := &ListNode{
		Val: num,
	}
	if s.Head == nil {
		s.Head = ele
	} else {
		ele.Next = s.Head
		s.Head = ele
	}
	s.Len++
}

输出

4
3
2
1

1
2
3
4