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("");}}