c语言写一算法，删除单链表中值为key的所有节点

在C语言中，删除单链表中所有值为`key`的节点，你可以编写如下的算法：

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

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

// 定义辅助函数用于查找并替换节点
Node* find_and_replace(Node* head, int key, int new_val) {
    if (head == NULL) return NULL;
    if (head->data == key) {
        head->data = new_val;
        return head;
    }
    Node* current = head->next;
    while (current != NULL) {
        if (current->data == key) {
            current->data = new_val;
        }
        current = current->next;
    }
    return head;
}

// 删除所有值为key的节点
void remove_all_nodes(Node** head, int key) {
    Node* temp = *head;
    Node* prev = NULL;

    while (temp != NULL) {
        if (temp->data == key) {
            if (prev == NULL) {  // 删除头节点的情况
                *head = temp->next;
            } else {
                prev->next = temp->next;
            }
            free(temp);
            temp = *head;  // 更新临时指针，继续遍历
        } else {
            prev = temp;
            temp = temp->next;
        }
    }
}

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

int main() {
    Node* list = create_linked_list();  // 这里假设你已经创建了一个链表
    printf("Original List: \n");
    print_list(list);

    remove_all_nodes(&list, 5);  // 删除值为5的节点

    printf("List after removing nodes with value 5: \n");
    print_list(list);

    return 0;
}