交换单个链接列表中的两个节点 [英] Swap two nodes in a singly linked list

问题描述

如何在pascal的单链接列表中切换两个节点?

How to switch two nodes in a singly linked list in pascal?

procedure List.switch(var n1,n2 : pNode);
var aux : pNode;
begin
  aux := n1;
  p1:= n2;
  n2 := aux;
end;

(这里pNode是列表中节点上的指针.)

(Here pNode is a pointer on a node in the list.)

节点的定义如下:

pNode = ^Node;
Node = record
    data : data;
    next : pNode;
end;

此代码无效.它要么不编译，要么说不能使用常量表达式的地址"，要么什么都不做.我想这与指针的工作方式有关...

This code doesn't work. It either doesn't compile, saying "Can't take the address of constant expressions", or just doesn't do anything. I guess it has to do with how pointers work...

我在此处找到了相关信息，但我没有找到读C.

I found relevant information here, but I don't read C.

感谢您的任何建议！

推荐答案

我认为这样会起作用:

function SwapNodes(first: pNode): pNode;
begin
  Result := first.next;
  first.next := Result.next;
  Result.next := first;
end;

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