如何获得两套在链表中共同的元素？ [英] How to get the common elements of the two sets in a Linked List?

问题描述

我有2链表，我想比较两个列表并打印出现在两个列表中的每个常见的元素。想尽一切无法得到它的工作。

 结构节点* calcIntersection（结构节点* headA，结构节点* headB）
{ 结构节点*链接1 = headA; 结构节点*链接2 = headB; 而（LINK1！= NULL）
 {  如果（link2-＆GT;价值== link2-＆GT;值）
     {      的printf（％d个，link1-＆GT;值）;     }    链接1 =链路1  - ＆GT; pNext; }  返回链接1;}
 

解决方案

您比较接近，在现实中你的方法应该是这样的：

 结构节点* calcIntersection（结构节点* headA，结构节点* headB）
{
    结构节点*链接1 = headA;    而（LINK1！= NULL）
    {
        结构节点*链接2 = headB;        而（LINK2！= NULL）
        {
            如果（link1-＆GT;价值== link2-＆GT;值）
            {
              的printf（％d个，link1-＆GT;值）;
            }            链接2 = link2-＆GT; pNext;
        }        链接1 = link1-＆GT; pNext;
    }   返回链接1;
}
 

我不太知道你正在返回那里，但你会被返回NULL大多数情况下，我不太清楚，如果这就是无论你想要什么。

I have got 2 linkedlist, I want to compare both lists and print every common element that appears in both list. tried everything can't get it to work.

struct Node *calcIntersection(struct Node *headA, struct Node *headB)
{

 struct Node * link1 = headA;

 struct Node * link2 = headB;

 while(link1 != NULL)
 {

  if (link2->value == link2->value)
     {

      printf("%d", link1->value);

     }      

    link1 = link1 -> pNext;

 }

  return link1;

}

解决方案

You are close, in reality your method should look like this:

struct Node *calcIntersection(struct Node *headA, struct Node *headB)
{
    struct Node *link1 = headA;

    while(link1 != NULL)
    {
        struct Node *link2 = headB;

        while (link2 != NULL)
        {
            if (link1->value == link2->value)
            {
              printf("%d", link1->value);
            }      

            link2 = link2->pNext;
        }

        link1 = link1->pNext;
    }

   return link1;
}

I'm not quite sure what you are returning there, but in most situations you will be returning NULL, I'm not quite sure if that's what you want however.

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