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-](/uploads/image/z/12185421-69-1.jpg?t=%E8%AE%BEp%E6%8C%87%E5%90%91%E5%8F%8C%E5%90%91%E9%93%BE%E8%A1%A8%E4%B8%AD%E6%9F%90%E7%BB%93%E7%82%B9%2Cs%E6%8C%87%E5%90%91%E5%BE%85%E6%8F%92%E5%85%A5%E7%9A%84%E5%80%BC%E4%B8%BAx%E7%9A%84%E6%96%B0%E7%BB%93%E7%82%B9%2C%E5%B0%86%2As%E6%8F%92%E5%85%A5%E5%88%B0%2Ap%E7%9A%84%E5%89%8D%E9%9D%A2%2C%E6%89%80%E9%9C%80%E8%A6%81%E7%9A%84%E6%93%8D%E4%BD%9C%E8%BF%87%E7%A8%8B%E6%98%AFA.%5Cx05%E2%91%A0+p-%3Eprior-%3Enext%3Ds%3B+%E2%91%A1+s-%3Eprior%3Dp-%3Eprior%3B+%E2%91%A2+s-%3Enext%3Dp%3B+%E2%91%A3+p-%3Eprior%3Ds%3BB.%5Cx05%E2%91%A0+s-%3Eprior%3Dp-%3Eprior%3B+%E2%91%A1+p-%3Eprior-)
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
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-
两个是一样的,①、②步两步都没有改变p->prior,所以说两步不分先后,没有差别~~~