02-线性结构2 Reversing Linked List

02-线性结构2   Reversing Linked List  (25分)

Given a constant $K$ and a singly linked list $L$, you are supposed to reverse the links of every $K$ elements on $L$. For example, given $L$ being 1→2→3→4→5→6, if $K=3$, then you must output 3→2→1→6→5→4; if $K=4$, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive $N$ (\le 10^5≤10​5​​) which is the total number of nodes, and a positive $K$ ($\le N$) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then $N$ lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Nextis the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

题目思路：关键是单链表的逆转，这是陈越老师给的思路，但是还有些测试点没有通过，需要再思考

#include
#include

#define MAXSIZE 100001 //100001=100000+1


typedef struct LNode *Ptr; 
struct LNode{
	int address;
	int data; 
	int next;
	Ptr link; //链表节点 
}list;

Ptr Reverse(Ptr head, int K,int L);
Ptr MakeList(int f_add, int next_add[],int data[],int L); 
void Attach(int a,int n_a,int data,Ptr *pRear);
void PrintList(Ptr P);


int main()
{
	int K = 0, L = 0; //K是需要翻转的节点数，L是所有节点数
	int link_num = 0;//链表节点数 
	int i;//用于计数 
	
	int address_tmp;//地址变量
	int first_add;//首地址 
	
	Ptr p,p_tmp,r_p;

	int Next[MAXSIZE] ;//用address当索引 
	int Data[MAXSIZE] ;//用address当索引 
	
	//读入首地址，数据数目，反转数目 
	scanf("%d %d %d",&first_add,&L,&K);
	//读入数据建立数据存储空间
	for(i = 0 ; i < L ; i++){ 
		scanf("%d",&address_tmp);//读入地址 
		scanf("%d %d",&Data[address_tmp],&Next[address_tmp]); 			
	}
 
 	//开始建立链表 
	p = MakeList(first_add,Next,Data,L);
	
	//反转链表 ******************需要修改*****************************
	if(K<=L){//K<=L才需要翻转 
		for(i = 0; i < (L/K); i++){
			r_p = Reverse(p,K,L);	
			p_tmp ->link = r_p;//多次翻转的考虑 
			p_tmp ->next =  r_p->address;
			for(int j = 0; j < 	K; j++ ){
	//使p指向下一段需要反转的子链表的头结点（第一个节点的前一个节点）  
				p_tmp = p_tmp->link;
			}
		} 
	}
	//************************************************************
	
	//输出链表
	PrintList(p_tmp); 
	 
	return 0;
} 
 
 
Ptr MakeList(int f_add, int next_add[],int data[],int L)
{
	Ptr P , Rear, t; 
	int t_L = L;
	int a,n_a,d,tmp_n_a;
	int i;
	
	P = (Ptr)malloc(sizeof(struct LNode));//链表头空节点
	P->link = NULL; 
	Rear = P;
	
	//增加首节点 
	a = f_add;
	d = data[a];
	n_a = next_add[a]; 
	Attach(a,n_a,d,&Rear); 
	t_L--; 
			
	while(t_L--){//其余节点增加 
		a = n_a;
		d = data[a];
		n_a = next_add[a];
		Attach(a,n_a,d,&Rear); 
	}
	//t = P;P=P->link;free(t);//删除临时生成头节点
	return P; 
}

void Attach(int a,int n_a,int data,Ptr *pRear)
{
	Ptr P;
	P = (Ptr)malloc(sizeof(struct LNode));
	
	P->address = a;
	P->data = data;
	P->next = n_a;
	P->link = NULL;
	
	(*pRear)->link = P;
	*pRear = P;	
}
 
Ptr Reverse(Ptr head, int K,int L)
{
	int cnt = 1;
	Ptr new_, old,tmp;
	//定义当前结点 new_，初始值为首元结点，new_ = head->link; 
	new_ = head->link;
	//定义当前结点的后继结点 old， old = new_->link; 
	old = new_->link;
	while(cnt < K){
		//定义新节点 tmp，它是 old的后继结点，tmp = old->link;
		tmp = old->link;
		//把old的后继指向new_, old->link = new_;  
		old->link =new_;
		//old的下一个地址是new_的地址  
		old->next = new_->address;
		//此时，old 实际上已经到了 new_ 前一位成为新的new_，
		//所以这个时候old 结点实际上成为新的 new_，new_ = old;  		
		new_ = old;
		//而新的 old 就是 tmp，old = tmp; 
		old = tmp;
		cnt ++; 
	} 
	///使反转后的最后一个节点指向下一段子链表的第一个节点  
	head->link->link = old;
	
	if(old != NULL){//对地址进行修改
		head->link->next = old -> address;	
	}
	else{
		head->link->next = -1;//正好完全翻转的情况 
	}
	return new_; 
}

void PrintList(Ptr P)
{
	while(P){
		if(P->next == -1) {//对最后一个节点的单独处理 
			printf("%.5d %d %d\n",P->address,P->data,P->next);
		}
		else{
			printf("%.5d %d %.5d\n",P->address,P->data,P->next);
		//格式输出，%.5意味着如果一个整数不足5位，输出时前面补0 如：22，输出：00022
		}		
		P = P->link;
	}	
}