1

我一直在研究反转链接列表的代码，但我似乎只在列表中打印出第一个元素。这是我的功能：

void reverseSqueue(Squeue squeue) {
    if (squeue -> first == NULL || squeue -> last == NULL){
        return;
    }
    Node* temp = NULL;
    Node* curr = squeue -> first;
    squeue -> first = squeue -> last;
    squeue -> last = curr;

    /*if (squeue -> first != squeue -> last){
      curr = curr -> next;
    }

    printf("%s\n", curr -> next -> value);
    printf("%s\n", curr -> value);
    printf("%s\n", curr -> prev -> value);
    printf("______________\n");
    printf("%s\n", squeue -> first -> value);
    Node* nde = squeue -> first;
    while (nde != NULL){
        printf("%s\n", nde -> value);
    nde = nde -> prev;
    }*/
    while (curr != NULL){
    /*if (curr != squeue -> first || curr != squeue -> last){
        if (curr -> next == squeue -> last){
            curr = NULL;
        }else{*/
        temp = curr;
        curr -> next = curr;
        curr = temp;
        curr = curr -> prev;
      //}
    //}else{
    //curr = NULL;
    //}
    }
}

我为混乱的代码道歉，我尝试运行多个测试并在这里注释掉它们；我之前的代码运行良好，但在 OClint 中给出了垃圾值警告，所以我尝试修复它们，现在我的代码似乎不再工作了。如果没有其他方法，我可以回到我的原始代码。非常感谢任何输入！

struct node {
  char *value;
  struct node *next;
  struct node *prev;
};
typedef struct node Node;

struct squeue {
   struct node* first;
   struct node* last;
};
typedef struct squeue * Squeue;
4

1 回答 1

0

使用 curr->next = curr 你是说当前节点的下一个节点是当前节点，也就是说，当前节点指向它自己，因为它是下一个节点。

Node *first = squeue->last; 
squeue->last = squeue->first; // the last node is now the first
squeue->first = first; // squeue->first = original order squeue->last

Node *curr = first; 
while (curr != NULL) { // loop begins with the original order last node,
                       // but new order first node

    // curr->next, curr->prev = curr->prev, curr->next
    Node *prev = curr->prev; // saves the curr's prev
    curr->prev = curr->next; // modifies curr's prev, now pointing to curr's next
    curr->next = prev; // curr's next now points to it's original prev

    curr = curr->next; // the next curr is the original order curr's prev,
                       // but the now order curr's next
}
于 2020-03-18T22:40:41.243 回答