Question

我有一个带有属性字符串和数字的输入我必须对与数字对应的字符串进行排序，如果两个字符串具有相同的对应数字，则字符串应按字典顺序排序。

例如输入

abc 2 嗨3 再见3

输出

再见3 嗨3 abc 2

如何在java中实现这一点。我应该选择哪种数据结构来解决这个问题？

Answer 1

您可以尝试这种方法

import java.io.*;
import java.util.*;

// required data structure
class A
{
String s;
int num;
}

class B
{
public static void main(String arg[])
{
Scanner sc=new Scanner(System.in);
String input=sc.nextLine();     //reading input

StringTokenizer st=new StringTokenizer(input);

Vector<A> v=new Vector<A>();        //for storing objects of class A

while(st.hasMoreTokens())       
{
A a=new A();        //creating objects
a.s=st.nextToken();
a.num=Integer.parseInt(st.nextToken());
v.add(a);
}

//sort

for(int i=0;i<v.size();i++)
{
    for(int j=0;j<v.size()-1;j++)
    {
        if(v.get(j).num<v.get(j+1).num)
        {
        A temp=v.get(j);
        v.set(j,v.get(j+1));
        v.set(j+1,temp);
        }
        else
        if(v.get(j).num==v.get(j+1).num)
        {
            if(v.get(j).s.compareTo(v.get(j+1).s)>0)
            {
            A temp=v.get(j);
            v.set(j,v.get(j+1));
            v.set(j+1,temp);
            }
        }
    }
}       //sort end

for(int i=0;i<v.size();i++)
System.out.print(v.get(i).s+" "+v.get(i).num+" ");
}
}

Answer 2

或者，您可以使用优先级队列并拥有一个存储类覆盖CompareTo。像这样：

/**
 * A container for data tuples, eg in "abc 2 hi 3 bye 3 q 2 a 1"
 * (abc 2) would be one DataValue Object
 * 
 * make it implement comparable and override compare to so that the objects with higher num 
 * values are considered a "higher priority"
 * @author Dragos
 *
 */
public class DataValue implements Comparable<DataValue> {

    private String str;
    private Integer num;

    public DataValue(String str, Integer num) {
        this.str = str;
        this.num = num;
    }


    @Override
    public int compareTo(DataValue o) {
        if (o.num > this.num)
            return 1;
        else
            return -1;
    }

    @Override
    public String toString() {
        return str + " " + num;
    }

    // GETTERS:

    public String getStr() {
        return str;
    }

    public Integer getNum() {
        return num;
    }

}

然后是主要课程：

import java.util.*;

public class SortingStrings {

    private static PriorityQueue <DataValue> pq = new PriorityQueue<DataValue>();

    public static void main (String [] args) {

        String inputString = "abc 2 hi 3 bye 3 q 2 a 1";
        // split the input into an array:
        String [] asArray = inputString.split(" ");

        // now load all the string and integer values into their own array lists:
        List<String> allSubStrings = new ArrayList<String> ();
        List<Integer> allSubNumbers = new ArrayList<Integer> ();


        for (int i = 0; i< asArray.length; i++) {

            // for even positions:
            if (i%2 == 0) 
                // we parse a string
                allSubStrings.add(asArray[i]);

            // for odd positions
            else 
                // we parse a number, and are ready to create a DataValue Object:
                allSubNumbers.add(Integer.parseInt(asArray[i]));

        }

        // now go through both array lists creating DataValue objects and adding them to the PriorityQueue:
        for (int i = 0; i < allSubStrings.size(); i++) {
            pq.add(new DataValue(allSubStrings.get(i),allSubNumbers.get(i)));
        }

        // now pop from the priority queue to get the resulting string:
        for (int i = 0; i < allSubNumbers.size(); i ++) {
            System.out.print(pq.remove() + " ");
        }


    }


}

Answer 3

import java.io.Console;
import java.util.Scanner;

public class s1 {

    public static void main(String[] args) {


        Scanner sc=new Scanner(System.in);
        String ss=sc.nextLine();
        String[] a;
        String temp;
        a = ss.split(" ");

        //sort 
        for (int i = 1; i < a.length; i += 2) {
            for (int j = i + 2; j < a.length; j += 2) {
                if (Integer.parseInt(a[i]) < Integer.parseInt(a[j])) {
                    //swapping if numbers were different
                    temp = a[i];
                    a[i] = a[j];
                    a[j] = temp;
                    temp = a[i - 1];
                    a[i - 1] = a[j - 1];
                    a[j - 1] = temp;

                } else if (Integer.parseInt(a[i]) == Integer.parseInt(a[j])) {
                    if (a[i - 1].compareTo(a[j - 1]) > 0) {
                        //swapping if numbers were same, then checking alphabets
                        temp = a[i];
                        a[i] = a[j];
                        a[j] = temp;
                        temp = a[i - 1];
                        a[i - 1] = a[j - 1];
                        a[j - 1] = temp;
                    }

                }
            }
        }

        for (int i = 0; i < a.length; i++)
            System.out.print(a[i]+" ");

    }

}

排序其对应的整数相同的值

3 个答案: