输入值从最低到最高的java排列

Question

请帮助我获取代码以安排最低到最高的输入值。以下是我的示例代码：

 System.out.print("Enter Number of Process: ");
 int p = Integer.parseInt(input.readLine());

 int [] arrival = new int[p];
 int [] size = new int[p];
 for(int i=0;i<p;i++){
 System.out.print("Arrival Time of Process #"+(i+1)+":");
 arrival[i]=Integer.parseInt(input.readLine());

预期的输出必须是：

输入进程数：4
进程#1的到达时间：100
进程#2的到达时间：110
进程#3的到达时间：102
进程#4的到达时间：101
P1：100
P4：101
P3：103
P2：110

如果你熟悉First Come First Serve Algorithm，那与此有关。

Answer 1

你可以使用方法 Arrays.sort(array[]);

    System.out.print("Enter Number of Process: ");
    int p = Integer.parseIntln(input.readLine());

    int [] arrival = new int[p];
    int [] size = new int[p];
    for(int i=0;i<p;i++) {
    System.out.println("Arrival Time of Process #"+(i+1)+":");
    arrival[i]=Integer.parseInt(input.readLine());
    }

    //You can make a copy of arrival[] and then sort it.
    //So arrival[] won't be sorted.
    size = arrival.clone();
    size = Arrays.sort(size);

    for(int i = 0; i < p; i++) {
    System.out.println("P"+(i+1)+": "+size[i]);
    }

Answer 2

使用Arrays.sort。

int[] is = {100, 110, 101, 102};
Arrays.sort(is);
for (int i : is) {
    System.out.println(i);
}

输出：

100
101
102
110

Answer 3

您似乎还想在进行排序时跟踪输入值的原始索引。我推荐你阅读的主题是sorting pairs of (value, index) compared by value。请尝试以下方法：

import java.io.IOException;
import java.util.Arrays;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    System.out.print("Enter Number of Process: ");
    int p = 0;

    try {
        p = Integer.parseInt(br.readLine());
    } catch (NumberFormatException nfe) {
        System.err.println("Invalid Format!");
    }

    Pair[] arrival = new Pair[p];
    for (int i = 0; i < p; i++) {
        System.out.print("Arrival Time of Process #" + (i + 1) + ":");
        try {

            arrival[i] = new Pair(i + 1, Integer.parseInt(br.readLine()));
        } catch (NumberFormatException nfe) {
            System.err.println("Invalid Format!");
        }
    }
    Arrays.sort(arrival);
    for (int i = 0; i < arrival.length; i++) {
        System.out.println("P" + arrival[i].index + ": " + arrival[i].value);
    }
}

// a little change was made to the original code
public static class Pair implements Comparable<Pair> {

    public final int index;
    public final int value;

    public Pair(int index, int value) {
        this.index = index;
        this.value = value;
    }

    @Override
    public int compareTo(Pair other) {
        return Integer.valueOf(this.value).compareTo(other.value);
    }
}

输出是：

Enter Number of Process: 4
Arrival Time of Process #1:100
Arrival Time of Process #2:110
Arrival Time of Process #3:103
Arrival Time of Process #4:101
P1: 100
P4: 101
P3: 103
P2: 110