golang实现链表逆置和删除元素

1、逆置

//链表逆置
type  Node struct{
	data int
	next *Node
}
func Reversenode(list *Node) *Node {
	if list==nil || list.next==nil || list.next.next==nil{
		return nil
	}
	p:=list.next
	list.next=nil
	var q *Node
	for p!=nil{
		q=p.next
		p.next=list.next
		list.next=p
		p=q
	}
	return list
}
func TestReversenode(t *testing.T) {

	head:=new(Node)
	l1:=new(Node)
	l1.data=1
	l2:=new(Node)
	l2.data=2
	l3:=new(Node)
	l3.data=3
	l4:=new(Node)
	l4.data=4
	l5:=new(Node)
	l5.data=5
	head.next=l1
	l1.next=l2
	l2.next=l3
	l3.next=l4
	l4.next=l5
	l5.next=nil

	for p:=head.next;p!=nil;p=p.next{
		fmt.Print(p.data)
	}

	Reversenode(head)
	fmt.Print("\n")
	for p:=head.next;p!=nil;p=p.next{
		fmt.Print(p.data)
	}


}

2、删除节点

//删除节点
func Findpri(list *Node, val int) *Node{        //找前驱
	if list==nil || list.next==nil{
		return nil
	}

	for p:=list;p.next!=nil;p=p.next{
		if p.next.data==val{
			return p
		}
	}
	return nil

}
func DeleteNode(list *Node, val int) bool{
	if list==nil || list.next==nil{
		return false
	}
	p:=Findpri(list,val)
	if p==nil{
		fmt.Println("not exist")
		return false
	}
	q:=p.next
	p.next=q.next
	return true
}
func TestDeleteNode(t *testing.T) {

	head:=new(Node)
	l1:=new(Node)
	l1.data=1
	l2:=new(Node)
	l2.data=2
	l3:=new(Node)
	l3.data=3
	l4:=new(Node)
	l4.data=4
	l5:=new(Node)
	l5.data=5
	head.next=l1
	l1.next=l2
	l2.next=l3
	l3.next=l4
	l4.next=l5
	l5.next=nil

	for p:=head.next;p!=nil;p=p.next{
		fmt.Print(p.data)
	}

	DeleteNode(head,1)
	fmt.Print("\n")
	for p:=head.next;p!=nil;p=p.next{
		fmt.Print(p.data)
	}


}
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转载自blog.csdn.net/YMY_mine/article/details/105105654