设p指向双向链表中某结点,s指向待插入的值为x的新结点,将s插入到p的前面,所需要的操作过程是A.\x05① p->prior->next=s; ② s->prior=p->prior; ③ s->next=p; ④ p->prior=s;B.\x05① s->prior=p->prior; ② p->prior-

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$设p指向双向链表中某结点,s指向待插入的值为x的新结点,将*s插入到*p的前面,所需要的操作过程是A.\x05① p->prior->next=s; ② s->prior=p->prior; ③ s->next=p; ④ p->prior=s;B.\x05① s->prior=p->prior; ② p->prior-$

设p指向双向链表中某结点,s指向待插入的值为x的新结点,将*s插入到*p的前面,所需要的操作过程是A.\x05① p->prior->next=s; ② s->prior=p->prior; ③ s->next=p; ④ p->prior=s;B.\x05① s->prior=p->prior; ② p->prior-
设p指向双向链表中某结点,s指向待插入的值为x的新结点,将*s插入到*p的前面,所需要的操作过程是
A.\x05① p->prior->next=s; ② s->prior=p->prior; ③ s->next=p; ④ p->prior=s;
B.\x05① s->prior=p->prior; ② p->prior->next=s; ③ s->next=p; ④ p->prior=s;
为什么选A不选B呢,我觉得两个都可以呀