C++ 删除无头链上所有指定值为x的节点。

C++ 删除无头链上所有指定值为x的节点。

#include<stdio.h>
#include<ctype.h>
#include<stdlib.h>
typedef struct app
{
    int data;
    struct app* next;
}APP;
int main()
{
    int n;
    int node;
    int x;
    while (scanf("%d", &n) != EOF)
    {
        APP* head = NULL, * p = NULL, * newnode = NULL;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &node);
            if (head == NULL)
            {
                head = (APP*)malloc(sizeof(APP));
                p = head;
                head->data = node;
                head->next = NULL;
            }
            else
            {
                newnode = (APP*)malloc(sizeof(APP));
                newnode->data = node;
                newnode->next = NULL;
                p->next = newnode;
                p = newnode;
            }
        }
        
        scanf("%d", &x);

        APP* pre = NULL;  //要删节点的前一个节点
        APP* curr = head;// curr当前
        while (curr) {
            if (curr->data == x) {
                
                //分两种情况处理
                if (pre == NULL) {//说明是链首删                    
                    APP* t = curr;
                    curr = curr->next;
                    free(t);
                    head=curr;
                    continue;
                }
                else {
                    pre->next = curr->next;
                    free(curr);
                    curr = pre->next;
                }
            }else{//不需要删的遍历下一条
                pre = curr;
                curr = curr->next;
            }             
        }



        p = head;
        while (p)
        {
            printf("%d  ", p->data);

            p = p->next;
        }
        puts("");

    }

}