编写程序删除单链表中所有值为x的节点--重制版_c语言实现单链表删除x

思路非常简单，就是查找出一个节点p满足值为x，将前驱节点q指向后继节点，然后将此节点删除即可。值得注意的是头节点head为此值时要分开讨论，即直接让头节点的后继节点成为新的头节点。

#include<stdio.h>
#include<math.h>
#include<malloc.h>
typedef struct node {
	int data;
	struct node* next;
}node;
node* Delete(node* head,int x) {
	node* p = head;
	node* q = NULL;
	while (p) {
		if (p->data == x&&p==head) {
			node* m = head;
			q = p;
			head = head->next;
			p = head;
			free(m);
		}
		else if (p->data == x) {
			q->next = p->next;
			node* m = p;
			p = p->next;
			free(m);
		}
		else {
			q = p;
			p = p->next;
		}
	}
	return head;
 
}
int main() {
	node* a = (node*)malloc(sizeof(node)); a->data = 1; 
	node* b = (node*)malloc(sizeof(node)); b->data = 2; 
	node* c = (node*)malloc(sizeof(node)); c->data = 3;
	node* d = (node*)malloc(sizeof(node)); d->data = 2;
	a->next = b; b->next = c; c->next = d; d->next = NULL;
	node* p = (node*)malloc(sizeof(node));
	p = Delete(a, 2);
	while (p) {
		printf("%d", p->data);
		p = p->next;
	}
	return 0;
}