逆置单链表C语言

  链表啊,我去年8月就认识你了.可是,居然到现在,我才写出来逆置单链表.拓展得远远不够啊.

  前一阵写过,但是没有实现.

  这次,昨晚构思完毕,刚才写出来了,而且还有链表本身.全下来30分钟,貌似比较快了,至少跟以前相比.

  好了,贴吧!当然了,这个ADT比较简洁.

/*10.2-7-2011-05-08-19.40.c -- 第十章第二节第七题*/#include <stdio.h>#include <stdlib.h>/*明显常量定义*/#define FALSE (0)#define TRUE (1)/*数据类型定义*/typedef int BOOL ;typedef int Item ;typedef struct node{Item item ;struct node * next ;} Node ;typedef Node * LinkedList ;/*接口函数声明*/void Initialize_L (LinkedList * const pl) ;BOOL IsEmpty_L (const LinkedList * const pl) ;BOOL Insert_L (LinkedList * const pl, const Item item) ;void Traversal_L (const LinkedList * const pl, void (* pfun) (const Item item)) ;void Release_L (LinkedList * const pl) ;/*接口函数定义*/void Initialize_L (LinkedList * const pl){*pl = NULL ;}BOOL IsEmpty_L (const LinkedList * const pl){if (NULL == *pl)return TRUE ;elsereturn FALSE ;}BOOL Insert_L (LinkedList * const pl, const Item item){Node * newNode ;newNode = (Node *) malloc (sizeof (Node)) ;if (NULL == newNode)return FALSE ;newNode -> item = item ;if (IsEmpty_L (pl)){newNode -> next = NULL ;*pl = newNode ;}else{newNode -> next = *pl ;*pl = newNode ;}return TRUE ;}void Traversal_L (const LinkedList * const pl, void (* pfun) (const Item item)){Node * scan ;for (scan = *pl; scan != NULL; scan = scan -> next)(* pfun) (scan -> item) ;}void Release_L (LinkedList * const pl){Node * scan, * temp ;scan = *pl ;while (scan != NULL){temp = scan ;scan = scan -> next ;free (temp) ;}*pl = NULL ;}/*主例程*/int mian (void) ;void printItem (const Item item) ;void reverse (LinkedList * const pl) ;int main (void){LinkedList list ;Item item ;Initialize_L (&list) ;item = 1 ;Insert_L (&list, item) ;item = 2 ;Insert_L (&list, item) ;item = 3 ;Insert_L (&list, item) ;item = 4 ;Insert_L (&list, item) ;item = 5 ;Insert_L (&list, item) ;Traversal_L (&list, printItem) ;reverse (&list) ;Traversal_L (&list, printItem) ;Release_L (&list) ;return 0 ;}void printItem (const Item item){printf ("%-3d/n", item) ;}void reverse (LinkedList * const pl){Node * current, * next, * temp ;/*If linkedlist is empty or only has one node.*/if (IsEmpty_L (pl) || NULL == (*pl) -> next)return ;current = *pl ;next = current -> next ;while (next -> next != NULL){temp = next -> next ;next -> next = current ;current = next ;next = temp ;}next -> next = current ;(*pl) -> next = NULL ;*pl = next ;}