英文:
How to update the next pointer in place in go list?
问题
我有一个来自列表的节点指针。我想用列表中的下一个节点更新值和下一个指针。这实际上就是在给定只有该指针的情况下进行原地删除。
例如：3 -> 5 -> 8 -> 9
要删除的节点：5（只给出5的访问权限，假设不知道前一个节点）
在这种情况下，可以将节点[8]的值和下一个指针复制到节点[5]。我有以下代码。它没有删除元素。如果我尝试使用'next'关键字访问下一个指针，它会抛出错误。
package main
import (
"container/list"
"fmt"
)
func main() {
l := list.New()
l.PushFront(4)
l.PushFront(5)
e4 := l.PushFront(7)
l.PushFront(6)
l.PushBack(9)
res := deleteNode(e4)
fmt.Println(res)
for e:=l.Front(); e!=nil;e=e.Next() {
fmt.Println(e.Value)
}
}
//ERROR
func deleteNode(node *list.Element) bool {
if node == nil || node.Next() == nil {
return false
}
var nextNode *list.Element
nextNode := node.next.(*list.Element)
node.Value = node.Next().Value.(int)
nextNode = nextNode.next.(*Element)
return true
}
有人可以帮我解决这个问题吗？
英文:
I have a node pointer from the list. I want update the value and next pointer with the next node in the list. This is nothing but deletion in place given access to only that pointer.
For example 3 -> 5 -> 8 -> 9
Node to be deleted : 5 (given access to only 5. Assuming previous node is not known)
In this case the value and next pointer of node[8] can be copied to node[5]. I have the following code. It is not removing the element. If I try to access the next pointer using 'next' keyword it is throwing error.
package main
import (
"container/list"
"fmt"
)
func main() {
l := list.New()
l.PushFront(4)
l.PushFront(5)
e4 := l.PushFront(7)
l.PushFront(6)
l.PushBack(9)
res := deleteNode(e4)
fmt.Println(res)
for e:=l.Front(); e!=nil;e=e.Next() {
fmt.Println(e.Value)
}
}
//ERROR
func deleteNode(node *list.Element) bool {
if node == nil || node.Next() == nil {
return false
}
var nextNode *list.Element
nextNode := node.next.(*list.Element)
node.Value = node.Next().Value.(int)
nextNode = nextNode.next.(*Element)
return true
}
Could anyone help me with this?
答案1
得分: 1
你得到的第一个错误是：
:= 左边没有新的变量
因为 nextNode 已经存在。问题出在这两行代码上：
var nextNode *list.Element
nextNode := node.next.(*list.Element)
在第一行中，你创建了变量 nextNode。在第二行中，你使用了短赋值 :=，它会创建一个新的变量（由左边指定），并将右边的值赋给它。只需省略第一行，你只需要其中一个：
nextNode := node.Next()
此外，你不能读取或更改节点的 next 指针，因为 Element 结构体中的 next 指针没有被导出（它以小写字母开头）。所以你想要实现的目标是无法完成的。但是你可以使用一个已定义的函数来实现：Remove(e *Element) interface{}
英文:
The (first) error you get is:
no new variables on left side of :=
Because nextNode already exists. The problem is with these 2 lines:
var nextNode *list.Element
nextNode := node.next.(*list.Element)
In the first line you create the variable nextNode. In the second line you use the short assignment := which creates a new variable (specified by the left side) and assigns to it the value on the right side. Just leave out the first line, you only need one of those:
nextNode := node.Next()
Moreover you cannot read or change the next pointer of a node because the next pointer is not exported in the Element struct (it starts with lowercased letter). So what you try to achieve cannot be done. But you have a function defined for this: Remove(e *Element) interface{}
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