带头结点的单链表删除

题目：在带头结点的单链表L中,删除所有值为X的节点，并释放其空间，假设值为X的节点不唯一
分析：
和上题相似，只是多了一个头结点。另我们可以采取直接遍历的方式进行删除，设置前驱结点，当前节点来保证不锻炼

代码：

struct Link {
 
	int data;
	struct Link *next;
};
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
void deleteX(Link *p,int delNum) {
	/*struct Link* q;//这是递归方法
	if (p == NULL) return;
	if (p->data==delNum) {
		q = p;
		p = p->next;
		free(q);
		deleteX(p,delNum);
	}
	else {
		deleteX(p->next, delNum);
	}*/
	//不采取递归，直接遍历
	struct Link *pre = p, *q = p->next,*r;
	while (q) {
		if (q->data==delNum) {
			r = q;//r指向待删除节点
			q = q->next;//
			pre->next = q;//删除节点
			free(r);//释放节点
		}
		else {
			pre = q;
			q = q->next;
			
		}
	}
}
int main() {
	//创建节点
	struct Link *head = (struct Link*)malloc(sizeof(struct Link));
	struct Link *q = (struct Link*)malloc(sizeof(struct Link));
	q = head;
	head->next = NULL;
	int n,data,delNum;
	printf("请输入节点个数：");
	scanf("%d",&n);
	for (int i = 0; i < n;i++) {
		printf("请输入第%d个节点值：",i+1);
		struct Link *p = (struct Link*)malloc(sizeof(struct Link));
		scanf("%d",&data);
		p->data = data;
		head->next = p;
		head = p;
	
	}
	head->next = NULL;//这里要将指针的next指向NULL，不然后面的判断会出问题，而且这也是应该养成的好习惯
	head = q;//head回到头结点
	printf("当前链表值为：");
	while (head->next) {
		printf("%d ",head->next->data);
		head = head->next;
	}
	printf("\n");
	printf("请输入要删除的值：");
	scanf("%d",&delNum);
	head = q;//head回到头结点
	deleteX(head,delNum);
	printf("删除后链表值为：");
	while (head->next) {
		printf("%d ", head->next->data);
		head = head->next;
	}
	return 0;
}

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