02-线性构造1. Reversing Linked List (25)
题目来源：
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 (<= 105) 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 33218Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1解法1： 有一个测试点通不过，原因是运行超时
import java.util.HashMap;
import java.util.Scanner;
public class Main{
static class Node
{
int value ;
String next ;
public Node(int value,String next)
{
this.value=value;
this.next=next;
}
public int getValue() {
return value;
}
public void setValue(int value) {
this.value = value;
}
public String getNext() {
return next;
}
public void setNext(String next) {
this.next = next;
}
}
/**
*（方法说明）
*@param 无
*@return 无
*@throws 无
*/
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
//头节点地址
String head =scanner.next();
//总共的节点数
int n = scanner.nextInt();
//转换因子 k
int k = scanner.nextInt();
//存储节点
HashMap<String,Node> hashMap = new HashMap<String,Node>();
for(int i=0;i<n;++i)
{
String key = scanner.next();
int value = scanner.nextInt();
String next = scanner.next();
Node node = new Node(value,next);
hashMap.put(key, node);
}
//存储要输出的4个address
String[] address =new String[k];
//address的下标
int index =0;
address[index++]=head;
Node node = hashMap.get(head);
//下一个地址的指针
String next = node.next;
//第一次输出
boolean firsline=false;
while(!next.equals("-1"))
{
if(index==k)
{
for(int i=(k-1);i>=0;--i)
{
if(i!=k-1 || firsline)
System.out.println(address[i]);
System.out.print(address[i]+" "+hashMap.get(address[i]).value+" ");
firsline = true;
}
index=0;
}
node = hashMap.get(next);
address[index++]= next;
next = node.next;
}
if(index>0)
{
if(index==k)
{
for(int i=(k-1);i>=0;--i)
{
if(i!=k-1 || firsline)
System.out.println(address[i]);
System.out.print(address[i]+" "+hashMap.get(address[i]).value+" ");
firsline = true;
}
index=0;
}
else
{
for(int i=0;i<index;++i)
{
System.out.println(address[i]);
System.out.print(address[i]+" "+hashMap.get(address[i]).value+" ");
}
}
}
System.out.println("-1");
}
}
解法2:参考http://blog.****.net/xtzmm1215/article/details/43195793
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
#define MAXN 100001
typedef struct{
int addr;
int data;
int next;
}Node;
Node nodes[MAXN];
vector<Node> list;
int main(){
int firstAdd, n, k;
scanf("%d%d%d", &firstAdd, &n, &k);
while(n--){
Node nn;
scanf("%d%d%d", &nn.addr, &nn.data, &nn.next);
nodes[nn.addr] = nn;
}
int address = firstAdd;
while(address != -1){
list.push_back(nodes[address]);
address = nodes[address].next;
}
int length = list.size();
int round = length/k;
for(int i = 1; i <= round; ++i){
int start = (i-1)*k;
int end = i*k;
reverse(list.begin() + start, list.begin() + end);
}
for(int i = 0; i < length-1; ++i){
printf("%05d %d %05d\n", list[i].addr, list[i].data, list[i+1].addr);
}
printf("%05d %d %d\n",list[length-1].addr, list[length-1].data, -1);
return 0;
}