golang 19题链表反转 超时

[IT-Talon]

代码如下

package main

import "fmt"

func main() {
	for {
		var n int
		_, err := fmt.Scan(&n)
		if err != nil {
			return
		}
		if n == 0 {
			fmt.Println("list is empty")
			continue
		}
		var head *Node
		dummyNode := &Node{next: head}
		temp := dummyNode
		for n > 0 {
			var num int
			_, err = fmt.Scan(&num)
			if err != nil {
				return
			}
			newNode := &Node{val: num}
			temp.next = newNode
			temp = temp.next
			n--
		}
		head = dummyNode.next
		show(head)
		show(head.reverse())
	}
}

func show(head *Node) {
	if head == nil {
		return
	}
	cur := head
	for cur != nil {
		fmt.Printf("%d ", cur.val)
		cur = cur.next
	}
	fmt.Println()
}

type Node struct {
	val  int
	next *Node
}

func (l *Node) reverse() *Node {
	if l == nil || l.next == nil {
		return l
	}
	var pre *Node
	cur := l
	for cur != nil {
		temp := cur.next
		cur.next = pre
		pre = cur
		cur = temp
	}
	return pre
}