两种方法求解 Reversing Linked List

问题描述：来自陈越

作者

CHEN, Yue

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 (<= 10  5 ) which is the total number of nodes, and a positive K (<=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   Next is 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

求解：

法一：链表的方法：

#include
#include
using namespace std;


typedef struct node
{
int Data;
int address;
int address_next;
struct node* next;
}Node;
Node *p;


int main()
{
int str,num,k;
int Adr[100001],NextAdr[100001],data[100001];
Node *p,*head,*tmp,*tmp1;
head=(Node*)malloc(sizeof(node)); //使用结点，链表务必记住要申请一个结点空间才行
tmp=(Node*)malloc(sizeof(node)); //使用结点，链表务必记住要申请一个结点空间才行

cin>>str>>num>>k;
tmp=head->next; // p is a empty head node
tmp->address=str;

///////////////////////////以下将输入的数据，地址转换为了一个链表／／／／／／／／／／／／／／／／／
for(int i=0;i{
cin>>Adr[i]>>data[i]>>NextAdr[i];
}
for(int i=0;i{
// cout<if(Adr[i]==tmp->address)
{
tmp->Data=data[i];
tmp->address_next=NextAdr[i];
}
}


while(tmp->address_next!=-1)
{
for(int i=0;i{
// cout<if(Adr[i]==tmp->address_next)
{
tmp1=(Node*)malloc(sizeof(node));
tmp1->address=Adr[i];
tmp1->Data=data[i];
tmp1->address_next=NextAdr[i];
tmp->next=tmp1;
tmp=tmp->next;
}
}
}


// while(head)
// {
// cout<Data<<" ";
// head=head->next;
// }
///////////////////////////以上将输入的数据，地址转换为了一个链表／／／／／／／／／／／／／／／／／
Node *new1,*old,*tmp2,*tmp3;   //链表反转
int n(1);
tmp3=head;
new1=head->next;
old=new1->next;
// tmp2=old->next;
while(n{
tmp2=old->next;
old->next=new1;
new1=old;
old=tmp2;
tmp2=tmp2->next;
n++;
}
head->next->next=old;
head->next=new1;


head=head->next;
while(head)
{
// cout<cout<address<<" "<Data<<" "<address_next<<" "<head=head->next;
}
return 0;
}

法二：采用数组，函数

#include
#include
using namespace std;


int main()
{
int list[100010];
// int address[100010];
int node[100010][2];
int num,str,k;
int address,data,next;
cin>>str>>num>>k;
for (int i = 0; i < num; ++i)
{
cin>>address>>data>>next;
node[address][0]=data;
node[address][1]=next;
// list[i]=address;
}
int m(0),n=str;

while(n!=-1)
{
list[m]=n;
m=m+1;
n=node[n][1];
}

for(int j=0;jcout<int j=0;
while(j+k{
reverse(list+j,list+j+k);
j=j+k;
}


for(int i=0;i{
cout<

}
}