使用C语言删除双向链表中的重复结点

可以按照以下步骤来删除双向链表中的重复结点：

1. 定义一个指向当前结点的指针和一个指向下一个结点的指针。
2. 遍历双向链表，对于每个结点，检查是否存在与之相同的结点。
3. 如果存在相同的结点，则删除该结点。删除结点时，需要修改前一个结点的 next 指针和后一个结点的 prev 指针。
4. 继续遍历链表，直到遍历完所有结点。

以下是一个示例代码：

#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int data;
    struct Node* prev;
    struct Node* next;
} Node;

Node* createNode(int val) {
    Node* node = (Node*)malloc(sizeof(Node));
    node->data = val;
    node->prev = NULL;
    node->next = NULL;
    return node;
}

void deleteDuplicates(Node* head) {
    Node* curr = head;
    while (curr != NULL) {
        Node* temp = curr->next;
        while (temp != NULL) {
            if (curr->data == temp->data) {
                Node* prev = temp->prev;
                Node* next = temp->next;
                prev->next = next;
                if (next != NULL) {
                    next->prev = prev;
                }
                free(temp);
                temp = next;
            } else {
                temp = temp->next;
            }
        }
        curr = curr->next;
    }
}

void printList(Node* head) {
    while (head != NULL) {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("\n");
}

int main() {
    Node* head = createNode(1);
    head->next = createNode(2);
    head->next->prev = head;
    head->next->next = createNode(3);
    head->next->next->prev = head->next;
    head->next->next->next = createNode(2);
    head->next->next->next->prev = head->next->next;
    head->next->next->next->next = createNode(4);
    head->next->next->next->next->prev = head->next->next->next;
    printf("Original List: ");
    printList(head);
    deleteDuplicates(head);
    printf("List after removing duplicates: ");
    printList(head);
    return 0;
}

输出结果为：

Original List: 1 2 3 2 4 
List after removing duplicates: 1 2 3 4