5

我已经尝试搜索与我的类似的问题，但没有找到太多帮助。

我有一个这种类型的结构的链接列表：

struct PCB {
    struct PCB *next;
    int reg1, reg2;
};

我首先创建了 10 个以这种方式链接在一起的 PCB 结构：

for(i=20;i<=30;i++) {
        curr = (struct PCB *)malloc(sizeof(struct PCB));
        curr->reg1 = i;
        curr->next  = head;
        head = curr;
    }

然后我需要再创建 20 个 PCB 结构，但它们的reg1值需要使用rand(). 我目前正在这样做：

for (j = 0;j<20;j++) {
        curr = (struct PCB *)malloc(sizeof(struct PCB));
        curr->reg1 = rand()%100;
        curr->next  = head;
        head = curr;
    }

但是，当将这些 PCB 结构插入到带有随机值的链表中时reg1，我需要将它们按顺序插入到链表中（插入排序）。仅在单链接链表中解决此问题的最佳方法是什么？谢谢

编辑：我现在正在跟踪第一个创建的结构，以便能够从一开始就遍历链表：

// create root struct to keep track of beginning of linked list
root = (struct PCB *)malloc(sizeof(struct PCB));
root->next = 0;  
root->reg1 = 20;

head = NULL;

// create first 10 structs with reg1 ranging from 20 to 30
for(i=21;i<=30;i++) {
    curr = (struct PCB *)malloc(sizeof(struct PCB));
    // link root to current struct if not yet linked
    if(root->next == 0){
        root->next = curr;
    }
    curr->reg1 = i;
    curr->next  = head;
    head = curr;
}

然后，当我创建另外 10 个需要插入排序的 PCB 结构时：

// create 20 more structs with random number as reg1 value
    for (j = 0;j<20;j++) {
        curr = (struct PCB *)malloc(sizeof(struct PCB));
        curr->reg1 = rand()%100;
        // get root for looping through whole linked list
        curr_two = root;
        while(curr_two) {
            original_next = curr_two->next;
            // check values against curr->reg1 to know where to insert
            if(curr_two->next->reg1 >= curr->reg1) {
                // make curr's 'next' value curr_two's original 'next' value
                curr->next = curr_two->next;
                // change current item's 'next' value to curr
                curr_two->next = curr;
            }
            else if(!curr_two->next) {
                curr->next = NULL;
                curr_two->next = curr;
            }
            // move to next struct in linked list
            curr_two = original_next;
        }
        head = curr;
    }

但这立即使我的程序崩溃了。

4

2 回答 2

6

“最好”的方法可能是为插入实现一个新功能。此函数将遍历列表，直到找到一个节点next值小于或等于您要插入的节点的节点，然后将新节点放在该节点之前next。

这个功能怎么样：

void insert(struct PCB **head, const int reg1, const int reg2)
{
    struct PCB *node = malloc(sizeof(struct PCB));
    node->reg1 = reg1;
    node->reg2 = reg2;
    node->next = NULL;

    if (*head == NULL)
    {
        /* Special case, list is empty */
        *head = node;
    }
    else if (reg1 < (*head)->reg1)
    {
        /* Special case, new node is less than the current head */
        node->next = *head;
        *head = node;
    }
    else
    {
        struct PCB *current = *head;

        /* Find the insertion point */
        while (current->next != NULL && reg1 < current->next->reg1)
            current = current->next;

        /* Insert after `current` */
        node->next = current->next;
        current->next = node;
    }
}

你可以这样称呼它：

insert(&root, rand() % 100, 0);
于 2013-04-11T22:36:15.740 回答
3

这是@Joachim 的简化版本：

void insert(struct PCB **head, const int reg1, const int reg2)
{
    struct PCB *new ;
        /* Find the insertion point */
    for (       ;*head; head = & (*head)->next)
    {
        if ((*head)->reg1 > reg1) break;
    }

    new = malloc(sizeof *new );
    new->reg1 = reg1;
    new->reg2 = reg2;
    new->next = *head;
   *head = new;
}

这个想法很简单：不需要任何特殊情况，在任何情况下：需要更改指针，这可能是根指针，尾指针，或 LL 中间的某个指针。在任何/每种情况下：

  • 新节点实际上窃取了这个指针：
  • 它使它指向自己
  • 它采用前一个值作为后继值（将其分配给它的->next指针。
于 2013-04-11T23:44:16.147 回答