反转链表

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 Lbeing 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 (≤) which is the total number of nodes, and a positive K (≤) 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
实现：

#include<iostream>

#include<stdio.h>

#include<algorithm> ///使用到reverse 翻转函数

using namespace std;

 

#define MAXSIZE 1000010 ///最大为五位数的地址

 

struct node ///使用顺序表存储data和下一地址next

{

int data;

int next;

}node[MAXSIZE];

 

int List[MAXSIZE]; ///存储可以连接上的顺序表

int main()

{

int First, n, k;

cin>>First>>n>>k; ///输入头地址 和 n，k；

int Address,Data,Next;

for(int i=0;i<n;i++)

{

cin>>Address>>Data>>Next;

node[Address].data=Data;

node[Address].next=Next;

}

 

int j=0; ///j用来存储能够首尾相连的节点数

int p=First; ///p指示当前结点

while(p!=-1)

{

List[j++]=p;

p=node[p].next;

}

int i=0;

while(i+k<=j) ///每k个节点做一次翻转

{

reverse(&List[i],&List[i+k]);

i=i+k;

}

for(i=0;i<j-1;i++)

printf("%05d %d %05d\n",List[i],node[List[i]].data,List[i+1]);

printf("%05d %d -1\n",List[i],node[List[i]].data);

return 0;

}