java - 模式识别

Question

所以我的代码已经完成，但现在我需要它来打印出所选数字是否相等!

我已经为 FALSE/NUMBERS ARE EQUAL 创建了一个循环，但它无法正常工作。希望你能帮忙。我的代码如下所示:

package patternrecognition;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Iterator;
import java.util.TreeMap;

public class PatternRecognition {

    public static void main(String[] args) throws Exception {



        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));


        int AntalNumre = -1;

        boolean Gyldignummer = false;
        while (Gyldignummer == false) {
            System.out.print("\n\nIndtast antal numre: ");

            String numre = reader.readLine().trim();

            Gyldignummer = validInteger(numre);

            if (Gyldignummer == false) {
                System.out.println("Indtast et gyldigt nummer");
            } else {

                AntalNumre = Integer.parseInt(numre);
            }

        }


        HashMap vaerdier = new HashMap();

        for (int i = 0; i < AntalNumre; i++) { //Studerendes nummer(i+1)

            boolean GyldigNummer2 = false;

            while (GyldigNummer2 == false) {
                System.out.print("\n\nIndtast en vaerdi for nummer " + (i + 1) + ": ");

                String vaerdi = reader.readLine().trim();
                int vaerdien = -1;

                GyldigNummer2 = validInteger(vaerdi);
                if (GyldigNummer2 == false) {
                    System.out.println("Indtast et gyldigt nummer");
                } else {

                    vaerdien = Integer.parseInt(vaerdi);
                }
                vaerdier.put(vaerdi, new Integer(vaerdien));
            }
        }



        TreeMap SorteretNummer = new TreeMap(vaerdier);

        Iterator nr = SorteretNummer.keySet().iterator();

        System.out.println("\n\n\n\n\n");
        System.out.println("Numre valgt:");
        System.out.println("------------");


        while (nr.hasNext()) {

            String navn = (String) nr.next();


            int numre = ((Integer) SorteretNummer.get(navn)).intValue();

            System.out.println("" + numre);

            if (numre != numre) {
                System.out.println("FALSE");
            }
            else {
                System.out.println("ALLE THE NUMBERS ARE EQUAL");
            }

        }
    }

    public static boolean validInteger(String nummer) {
        boolean validInteger = false;


        try {
            Integer.parseInt(nummer);
            validInteger = true;
        } catch (NumberFormatException nfe) {
            validInteger = false;
        }
        return validInteger;

    }
}

Answer 1

嗯，numre != numre必然是false当然除非numre碰巧在另一个线程中被更改(或者是 NaN ，但这是一个不同的故事)。

也许，您可能想比较数字对？或者，也许您想要一个数据结构来保存您已经见过的数字(例如 HashSet 或 BitSet )？

示例:当用户输入数字时，将其保存到 HashSet<Integer> 。一旦完成，添加

Set<Integer> set = new hashSet<Integer>();

// ...as we're getting numbers from user

   set.add(numre);

// ...

if (set.size() == 1) 
    System.out.println("OMG they're all the same! " + set);