（程序员面试题）单链表的反转

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只用一个额外的节点实现单链表的反转

testcase如下：

#include <stdio.h>#include <stdlib.h>typedef struct num {    int data;    struct num *next;} *num_ptr;void print_num(num_ptr head) {      num_ptr tmp = head;      while (tmp->next) {          printf("%d\n", tmp->next->data);          tmp = tmp->next;      }      printf("\n");  }  num_ptr reverse_link(num_ptr head) {    num_ptr priv = head->next;    num_ptr cur = head->next->next;    num_ptr tmp = (num_ptr)malloc(sizeof(struct num));    tmp->next = cur->next;        priv->next = NULL;    while (cur->next) {        // start reverse        cur->next = priv;        // priv change to next        priv = cur;        // cur change to next, this is why need tmp node        cur = tmp->next;        // tmp still save cur next        tmp->next = cur->next;    }    // change last node next null to priv    cur->next = priv;    head->next = cur;    return head;}int main(){    num_ptr head, one, two, three, four;    head = (num_ptr)malloc(sizeof(struct num));    one = (num_ptr)malloc(sizeof(struct num));    two = (num_ptr)malloc(sizeof(struct num));    three = (num_ptr)malloc(sizeof(struct num));    four = (num_ptr)malloc(sizeof(struct num));    one->data = 1;    two->data = 2;    three->data = 3;    four->data = 4;    head->next = one;    one->next = two;    two->next = three;    three->next = four;    four->next = NULL;    print_num(head);    num_ptr new_head = reverse_link(head);    print_num(new_head);    return 0;}

这个问题的关键就在于那一个额外的节点，因为当链表做完反转之后就无法指向链表的下一个节点了，这个时候用额外的节点去记住反转的节点的下一个节点，这样子就可以继续进行下一轮反转了

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