【参考答案】从第一个元素开始，ps指向他，将他（ps）指向头节点(ps->next = head) ，将ps设为头节点（head = ps; ）操作下一个元素（ps= pe->next;）等于是依次将每个元素翻到原头节点前面。
void reverse(test* head){test* pe = head;test* ps = head->next;while(ps){ pe->next = ps->next;ps->next = head;head = ps;ps = pe->next;}}
