单链表的就地逆置 （单链表带头结点）（某公司校园招聘面试试题）

      面试某公司，被问此题，由于紧张，程序有误，甚为遗憾！下面给出正确的程序：

#include<iostream>using namespace std;typedef struct node{int data;struct node *next;}Node, *List;List createList(){Node *head, *p1, *p2;p1 = p2 = head = new Node;int num;cin >> num;while(-1 != num){p1 = new Node; p1->data = num;p2->next = p1;p2 = p1;cin >> num;}p2->next = NULL;return head;}void printList(List p){while(NULL != p->next){cout << p->next->data << " ";p = p->next;}cout << endl;}void releaseList(List p){if(NULL == p->next)delete p;else{releaseList(p->next);delete p;}}List reverseList(List p){//让p1和p2都指向第一个结点Node *p1 = p->next;Node *p2 = p1;p->next = NULL;//分离p与p1while(NULL != p1){p2 = p1;p1 = p1->next; //在"摘下" p2时，千万记得先进行p1 = p1->next;//倒插法重建链表p2->next = p->next;p->next = p2;}return p;}int main(){List head = createList();printList(head);reverseList(head);printList(head);releaseList(head);return 0;}