问题：将一个链表反转写出算法。
回答：方法一：
#include<stdio.h>
#include<stdlib.h>
/* 链表节点 */
struct node
{
int data;
struct node* next;
};
/* 反转单链表. 分别用3个指针，指向前一个，当前，下一个 */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* 添加数据。 头部插入 */
void push(struct node** head_ref, int new_data)
{
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
/* 打印链表 */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf("%d ", temp->data);
temp = temp->next;
}
}
int main()
{
struct node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
push(&head, 60);
push(&head, 100);
printList(head);
reverse(&head);
printf("
Reversed Linked list
");
printList(head);
}
方法二：
/* 使用递归的方法 */
static struct node * reverseRecall(struct node* head){
//最后一个节点会返回 作为头部
if(NULL == head || head->next == NULL) return head;
//head->next 表示剩下的部分
struct node * newHead = reverseRecall(head->next);
head->next->next = head; //颠倒指针
head->next = NULL;//原来的头节点 next 应该为空
return newHead;
}