java - 如果这两个数组在同一位置具有相同的数字，如何显示？

Question

所以这是我的问题:我有两个数组，我想计算两个数组上完全相同的位置 [i] 有多少个值，以及两个数组上有多少个值相同但 [i] 不同

我尝试了使用 for 和数组大小的通常循环，但它显示了与预期相差甚远的奇怪值

package stackOverflow;

import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.Arrays;

import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JTextField;

public class mainStack extends JFrame implements ActionListener {

    JButton jbr,jbv,jbb,jbo,jbn,jbj; 
    JTextField l11b;
    String a;

    int tabRef[]= {0,1,2,3};
    int correctAndSameCase=0;
    int correctButDiffCase=0;
        mainStack(){

            this.setLayout(null);
            jbr = new JButton("Rouge");
            jbr.setBounds(0,80,85,30);
            add(jbr);

            jbv = new JButton("Vert");
            jbv.setBounds(125, 80, 85, 30);
            add(jbv);

            jbb = new JButton("Bleu");
            jbb.setBounds(0, 120, 85, 30);
            add(jbb);

            jbj = new JButton("Jaune");
            jbj.setBounds(125, 120, 85, 30);
            add(jbj);

            jbo = new JButton("Orange");
            jbo.setBounds(0, 160, 85,30);
            add(jbo);

            jbn = new JButton("Noir");
            jbn.setBounds(125,160, 85,30);
            add(jbn);

            jbr.addActionListener(this);
            jbv.addActionListener(this);
            jbb.addActionListener(this);
            jbj.addActionListener(this);
            jbo.addActionListener(this);
            jbn.addActionListener(this);

            setLayout(null);
            setSize(800,800);
            setVisible(true);
        }
        private int index = 0;
        private int p=0;
        private int tabAnswer[][] = new int[3][4];
        private int i;


        public void actionPerformed(ActionEvent e) {
            if (e.getSource().equals(jbr)) {
                tabAnswer[p][index] = 0;
            } else if (e.getSource().equals(jbv)) {
                tabAnswer[p][index] = 1;
            } else if (e.getSource().equals(jbj)) {
                tabAnswer[p][index] = 2;
            } else if (e.getSource().equals(jbb)) {
                tabAnswer[p][index] = 3;
            } else if (e.getSource().equals(jbo)) {
                tabAnswer[p][index] = 4;
            } else if (e.getSource().equals(jbn)) {
                tabAnswer[p][index] = 5;
            }
            index++;
            if (index >= tabAnswer[p].length) {
                System.out.println(Arrays.toString(tabAnswer[p]));

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

                    if(tabAnswer[p][i]==tabRef[i]) {

                        correctAndSameCase++;
                    }else if(tabAnswer[p][i]==tabRef[0] & tabAnswer[p][i]!=tabRef[i]  ||  tabAnswer[p][i]==tabRef[1] & tabAnswer[p][i]!=tabRef[i] || tabAnswer[p][i]==tabRef[2]& tabAnswer[p][i]!=tabRef[i] ||tabAnswer[p][i]==tabRef[3] & tabAnswer[p][i]!=tabRef[i]) {

                        correctButDiffCase++;
                    }

                }
                index = 0;
                p++;
                System.out.println(correctAndSameCase+" number are on the same case on both arrays");
                System.out.println(correctButDiffCase+" number are on different case on both arrays");
                if(p>=3) {
                    p=0;
                }
                correctAndSameCase=0;
                correctButDiffCase=0;

            }
        }
    public static void main(String[] args) {
        mainStack t= new mainStack();
}

这里的 tabAnswer {5,4,3,2} CorrectAndSameCase 应该是 0 ， CorrectButDiffCase 应该是 2 但它给出我 0 和 1 作为答案。

编辑:我可以看到问题在于将值设置为 CorrectButDiffCase 但我不知道如何修复它

Answer 1

这里有一个简短的代码片段，用于计算这两个值。但我不知道，是否要计算一些特殊情况(请参阅 TODO)。

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

    List<Integer> tabRef =  Arrays.asList(10, 5, 3, 5, 2, 6);
    List<Integer> tabRef2 = Arrays.asList(12, 8, 3, 5, 6, 2);

    int matchesOnSameIndex = 0;
    int matchesButDifferentIndex = 0;

    for (int i = 0; i < tabRef.size(); i++) {
        //compare on same index, if other list has element on this index
        if (tabRef2.size() > i) {
            if (tabRef.get(i) == tabRef2.get(i)) {
                //same element on same index
                matchesOnSameIndex++;
            }
        }

        //check if value exists in other list
        if (tabRef2.contains(tabRef.get(i))) {
            //if yes, check if it is not at same position
            if (tabRef2.size() > i) {
                if (tabRef2.get(i) != tabRef.get(i)) {
                    //not same index
                    //TODO must count multiple times, if element exists multiple times?
                    matchesButDifferentIndex++;
                }else {
                    //TODO must count, if also exists on other index?
                }
            }else {
                //same position not existing, so must be on other position
                matchesButDifferentIndex++;
            }
        }
    }

    System.out.println("matchesOnSameIndex: "+matchesOnSameIndex);
    System.out.println("matchesButDifferentIndex: "+matchesButDifferentIndex);

}