链表从指定的节点前插入新的节点

1.第一种情况：

在头节点插入时，head和p都指在1，new ->next = head;return new;返回new节点地址，由打印函数打印即可成功插入

2.第二种情况：

2.1当我们要在第2个节点插入时，p和head都是指向1节点的，我们得用while循环判断p->next !=NULL;（是否等于空）

然后用if()函数把判断p->next->data 与我们要在所在的节点的data值相比是否相等；否则：printf("no this data %d",data);返回head(头节点)；

如果相等则：new->next = p->next ; p ->next = new; 插入成功，返回head，

3.第三种情况：

当我们想要改变的节点都不在1，2节点，而头节点又在1节点位置时；我们可以在以上两种情况的基础上用

p = p ->next;

从头遍历节点！

为什么返回head?

because head list never change!!!

#include<stdio.h>
struct Test
{
        int data;
        struct Test *next;
};

void printLink(struct Test *head)
{
        struct Test *point;
        point = head;
        printf("use Link to print num:\n");
        while(point !=NULL){
                printf("%d ",point->data);
                point = point->next;
        }
        putchar('\n');
}

int getLinkTotalnums(struct Test *head)
{
        int count = 0;
        while(head != NULL){
                count++;
                head = head ->next;
        }
        return count;
}

struct Test *insertFromFor(struct Test *head,int data,struct Test *new)
{
        struct Test *point;
        point = head;
        if(point->data == data){
                new ->next =head;
                return new;
        }
        while(point ->next !=NULL){
                if(point->next->data == data){
                        new ->next = point ->next;
                        point ->next = new;
                        printf("insert is ok!\n");
                        return head;
                }
 point = point ->next;
        }
        printf("no have this data %d\n",data);
        return head;
}

struct Test *insertFromHead(struct Test *head,int data,struct Test *new2)
{
        struct Test *p;
        p = head;
        if(p->data == data){
                new2->next = head;
                return new2;
        }
}

int main()
{
        int i;
        int arr[] = {1,2,3,4,5,6,7,8,9,10};
        struct Test *head;

        for(i =0 ;i<sizeof(arr)/sizeof(arr[0]);i++){
                printf("%d ",arr[i]);
        }
        putchar('\n');
        struct Test t1 = {1,NULL};
        struct Test t2 = {2,NULL};
        struct Test t3 = {3,NULL};
        struct Test t4 = {4,NULL};
        struct Test t5 = {5,NULL};

        struct Test new2 = {666,NULL};
        struct Test new3 = {999,NULL};
        head = &t1;
        t1.next = &t2;
        t2.next = &t3;
        t3.next = &t4;
        t4.next = &t5;
        printLink(head);

        head = insertFromHead(head,1,&new2);
        printLink(head);

        head = insertFromFor(head,4,&new3);
        printLink(head);

        return 0;
}

实验效果图：