双向链表中的结点删除问题

Question

给出一个数x，从双向链表中删除所有元素x的结点

Answer 1

大体思路是：
（1）遍历链表
（2）判断当前节点的data是否等于x
（3）如果等于x，ppre记录上一个节点，nnext记录下一个节点，
ppre->next = nnext; nnext->pre = ppre;
删除当前节点的内存
继续判断下一个节点。
（4）如果不等于，继续遍历下一个节点

代码如下：

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

struct stnode 
{
    int data;
    struct stnode* pre;
    struct stnode* next;
};


//创建双向链表
struct stnode* create()
{
    int i;
    struct stnode* head,*p,*t;
    head = (struct stnode*)malloc(sizeof(struct stnode));
    head->pre = NULL;
    head->next = NULL;

    t = head;
    //创建5个节点
    for (i=0;i<5;i++)
    {
        p = (struct stnode*)malloc(sizeof(struct stnode));
        p->pre = NULL;
        p->next = NULL;
        printf("请输入1个整数：");
        scanf("%d",&p->data);
        t->next = p;
        p->pre = t;
        t = p;
    }
    return head;
}


//删除值为x的节点
struct stnode* del(struct stnode* head,int x)
{
    struct stnode *p,*ppre,*nnext;
    p = head->next;
    while (p)
    {
        if(p->data == x)
        {
            ppre = p->pre;
            nnext = p->next;
            ppre->next = nnext;
            if(nnext)
                nnext->pre = ppre;
            free(p);
            p = nnext;
        }else
            p = p->next;
    }
    return head;
}

//xians
void show(struct stnode* head)
{
    struct stnode* p = head->next;
    while(p)
    {
        printf("%d ",p->data);
        p = p->next;
    }
}

//free
void Release(struct stnode* head)
{
    struct stnode* p;
    while(head)
    {
        p = head->next;
        free(head);
        head = p;
    }
}

int main()
{
    int x;
    struct stnode* head = create();
    show(head);
    printf("请输入需要删除的值：");
    scanf("%d",&x);
    head = del(head,x);
    //显示
    show(head);
    //释放空间
    Release(head);
    return 0;
}