【算法每日一练】PAT甲级1052 java

A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive N (<105) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by −1.

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

Address Key Next

where Address is the address of the node in memory, Key is an integer in [−105,105], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.

Output Specification:

For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.

Sample Input:

5 00001
11111 100 -1
00001 0 22222
33333 100000 11111
12345 -1 33333
22222 1000 12345

      
    

Sample Output:

5 12345
12345 -1 00001
00001 0 11111
11111 100 22222
22222 1000 33333
33333 100000 -1

      
    

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题意

给出N个结点的地址address，数据域data以及指针域next,然后给出链表的首地址，把这个链表上的节点按data值从小到大除数

初始设置所有节点均为无效节点，然后根据首地址进行遍历，遍历到的标识记为true

对节点进行排序

代码

package com.zixin.algorithm;

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class PATA1052 {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();// 数量
		int head = sc.nextInt();// 头结点
		Node52[] arr = new Node52[100005];
		for (int i = 0; i < 100005; i++) {// 这个地方需要优化
			Node52 node = new Node52();
			node.flag = false;
			arr[i] = node;
		}
		for (int i = 0; i < n; i++) {
			Node52 node = new Node52();
			int address = sc.nextInt();
			node.address = address;
			node.data = sc.nextInt();
			node.next = sc.nextInt();
			arr[address] = node;
		}
		sc.close();
		int count = 0;
		int p = head;
		while (p != -1) {
			arr[p].flag = true;
			count++;
			p = arr[p].next;
		}
		if (count == 0) {// 特判 链表中没有节点 输出 0 -1
			System.out.println("0 -1");
		} else {
			Arrays.parallelSort(arr, new Compa52());// 有效节点进行排序
			System.out.printf("%d %05d\n", count, arr[0].address);// 防止-1被格式化 上面特殊处理 输出节点信息 下一个节点不是节点类的next
																	// 而是下一个节点的address
			for (int i = 0; i < count; i++) {
				if (i != count - 1) {
					System.out.printf("%05d %d %05d\n", arr[i].address, arr[i].data, arr[i + 1].address);
				} else {
					System.out.printf("%05d %d -1\n", arr[i].address, arr[i].data);
				}
			}
		}
	}

}

class Node52 {
	int address;
	int data;
	int next;
	boolean flag;

}

class Compa52 implements Comparator<Node52> {

	@Override
	public int compare(Node52 o1, Node52 o2) {
		if (!o1.flag || !o2.flag) {
			if (!o1.flag && o2.flag) {
				return 1;
			} else if (o1.flag && !o2.flag) {
				return -1;
			} else {
				return 0;
			}
		} else {
			return o1.data - o2.data;
		}
	}

}

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