编写程序删除单链表中所有值为x的节点

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编写程序删除单链表中所有值为x的节点

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;
}