提问，输入一个数，在单链表中删除第一个与数相同的元素

Question

报错图

运行图

相关代码

// Delete the first node in the list containing the value `value`.
// The deleted node is freed.
// If no node contains `value`, the list is not changed.
// The head of the list is returned.
//删除包含值“value”的列表中的第一个节点。
//已删除的节点将被释放。
//如果没有节点包含“value”，则列表不会更改。
//返回head。
//
struct node *delete_contains(int value, struct node *head) {

    struct node *p;
    p = head;
    p->data = value;
    while (head != NULL && head->next != NULL) {
        if (head->data == p->data) {

            p->next = p->next->next;
            free(p);
        } else {
            head = head->next;
        }
    }

    return head;

}

//其他不需要改的代码

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

struct node {
    struct node *next;
    int          data;
};

struct node *delete_contains(int value, struct node *head);
struct node *strings_to_list(int len, char *strings[]);
void print_list(struct node *head);

int main(int argc, char *argv[]) {
    int value;
    scanf("%d", &value);
    // create linked list from command line arguments
    struct node *head = strings_to_list(argc - 1, &argv[1]);

    struct node *new_head = delete_contains(value, head);
    print_list(new_head);

    return 0;
}




struct node *strings_to_list(int len, char *strings[]) {
    struct node *head = NULL;
    int i = len - 1;
    while (i >= 0) {
        struct node *n = malloc(sizeof (struct node));
        assert(n != NULL);
        n->next = head;
        n->data = atoi(strings[i]);
        head = n;
        i -= 1;
    }   
    return head;
}

void print_list(struct node *head) {
    printf("[");    
    struct node *n = head;
    while (n != NULL) {
        // If you're getting an error here,
        // you have returned an invalid list
        printf("%d", n->data);
        if (n->next != NULL) {
            printf(", ");
        }
        n = n->next;
    }
    printf("]\n");
}

Answer 1

struct node *delete_contains(int value, struct node *head) {
 
    struct node *p,*t;
    p = head;
    //p->data = value;
    while (p != NULL && p->next != NULL) {        
        if (p->next->data == value) {
            t=p->next;
            p->next = p->next->next;
            free(t);//此时如果是p，则链表断了。
        } else {
            p = p->next;
        }
    } 
    return head;//原函数中head已经是n个next之后的节点了 
}