如何为链表实现最小和最大节点功能？ [英] How to implement minimun and maximum node function for a linked list?

问题描述

void max ()
{
struct node *max = (struct node*)malloc(sizeof(struct node));
struct node *min = (struct node*)malloc(sizeof(struct node));
struct node *temp = head->next;

if (head ==  NULL){ printf("list is empty\n");}
max-> data = temp-> data;
min-> data = temp-> data;
while(temp->next!=NULL)
{
if(temp->data > max->data){max->data = temp->data;}
else if (temp->data < min->data){min->data = temp->data;}
temp=temp->next;
}
printf(" min = %d // max = %d",min->data, max->data);

}

我尝试过：



这是一个打印链表最小和最大值的函数



如果链接列表i / p ______：1 2 3 4 5 6

_______________ o / p是min = 2 / max = 5



是不考虑第一个或最后一个节点，如何解决？

What I have tried:

this is a function to print the min and max of a linked list

if the linkedlist i/p______ : 1 2 3 4 5 6
_______________o/p is min = 2 / max = 5

is does not take first or last nodes into account, how to fix that?

推荐答案

不要将新节点分配给Max或Min ，并且不要从第二个元素开始！

相反，将最初的Max和Min指针设置为列表头：

Don't assign new Nodes to Max or Min, and don't start with the second element!
Instead, set the initial Max and Min pointers to the list head:

struct node *max = head;
struct node *min = head;
struct node *temp = head;

然后在循环中使用它们：

Then use them in your loop:

while (temp != NULL)
   {
   ...
   temp = temp->next;
   }

在循环中，将临时值与min和max进行比较，如果找到新值，则将相应的指针设置为temp。 />


这样，你就不会浪费内存 - 你不会释放内存，导致内存泄漏 - 而且你会覆盖列表中的所有元素。

如果列表包含一个元素，则max和min都是相同的：唯一的节点。



尝试一下：当你想到它时它很明显it！

Inside the loop, compare the temp value with min and max, and if you find a new value, set the appropriate pointer to temp.

That way, you aren't wasting memory - which you don't free, causing memory leaks - and you cover all elements in the list.
If the list contains one element, the max and min are both the same: the one and only Node.

Try it: it's pretty obvious when you think about it!

建议：正确使用你的代码，它有助于阅读。

Advice: intend your code properly, it help reading.

void max ()
{
	struct node *max = (struct node*)malloc(sizeof(struct node));
	struct node *min = (struct node*)malloc(sizeof(struct node));
	struct node *temp = head->next;

	if (head ==  NULL){ printf("list is empty\n");}
	max-> data = temp-> data;
	min-> data = temp-> data;
	while(temp->next!=NULL)
	{
		if(temp->data > max->data){
			max->data = temp->data;}
		else if (temp->data < min->data){
			min->data = temp->data;}
		temp=temp->next;
	}
	printf(" min = %d // max = %d",min->data, max->data);

}

当你不明白你的代码在做什么或为什么它做它做的事情，答案是调试器。

使用调试器查看代码正在做什么。只需设置断点并查看代码执行情况，调试器允许您逐行执行第1行并在执行时检查变量，这是一个令人难以置信的学习工具。



调试器 - 维基百科，免费的百科全书 [ ^ ]

掌握Visual Studio 2010中的调试 - 初学者指南 [ ^ ]

使用Visual Studio 2010进行基本调试 - YouTube [ ^ ]



调试器在这里向您展示您的代码在做什么你的任务是与它应该做的事情进行比较。

调试器中没有魔法，它没有发现错误，它只是帮助你。当代码没有达到预期的效果时，你就会接近一个错误。

When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.

Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.

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