Question

我正在开发一个程序，我们比较两个字符串，并将一个字符串中不存在的字符添加到另一个字符串，反之亦然。例如：String 1- aabccd 字符串2- acccdd 输出应该是 - 要添加到字符串1- cd的字符要添加到String 2- ab的字符当角色不重复时，我能够实现它，例如：String 1-芒果弦乐2盎司输出是 - 要添加到字符串1- l的字符要添加到字符串2- m的字符但是当角色重复时却无法得到它。

这是我写的代码 -

import java.util.Scanner;

public class Test {

public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);

        String S1, S2;

        System.out.println("Enter String 1: ");
        S1 = user_input.next();

        System.out.println("Enter String 2: ");
        S2 = user_input.next();

        user_input.close();

        char[] S1Array = S1.toLowerCase().toCharArray();
        char[] S2Array = S2.toLowerCase().toCharArray();
        charAddition(S1Array, S2Array);
        charAdditionReverse(S2Array,S1Array);
}
private static void charAddition(char[] n, char[] S1Array) {
for (char n1 : n) {
            if (!isPresent(n1, S1Array)) {
                System.out.println("character to be added to S2Array
is: " + n1);
            }
        }
}
private static void charAdditionReverse(char[] n, char[] S1Array) {
for (char n1 : n) {
            if (!isPresent(n1, S1Array)) {
                System.out.println("character to be added to S1Array
is: " + n1);


            }
        }
}


    private static boolean isPresent(char n, char[] S1Array) {
        for (char i : S1Array) {
            if (n == i) {
                return true;
            }
        }
        return false;
    }
}

Answer 1

以下是您的解决方案：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;
import java.util.Scanner;

public class Test {

    public static void main(String[] args) {
        Scanner user_input = new Scanner(System.in);

        String S1, S2;

        System.out.println("Enter String 1: ");
        S1 = user_input.next();

        System.out.println("Enter String 2: ");
        S2 = user_input.next();

        user_input.close();

        char[] S1Array = S1.toLowerCase().toCharArray();
        char[] S2Array = S2.toLowerCase().toCharArray();
        charAddition(S1Array, S2Array,false);
        charAddition(S2Array,S1Array,true);
    }
    private static void charAddition(char[] n, char[] S1Array,boolean isreverse) 
    {

        List<char[]> asList = Arrays.asList(S1Array); // because this DOES compile.

        List<Character> wordlist = new ArrayList<Character>();
        for (char c : S1Array) {
            wordlist.add(c);
        }

        for (char n1 : n) 
        {
            if (!isPresent(n1, wordlist)) 
            {
                if(!isreverse)
                {
                    System.out.println("character to be added to S2Arrayis: " + n1);

                }else
                {
                    System.out.println("character to be added to S1Arrayis: " + n1);
                }
            }
        }
    }


    private static boolean isPresent(char n, List<Character> S1Array) 
    {
        boolean flag ;
        for (Iterator<Character> it = S1Array.iterator(); it.hasNext();) 
        {
            char i = it.next();
            if (n == i) 
            {
                it.remove();
                return true;
            }
        }
        return false;
    }
}

要解决上述问题，您必须检查特定字符重复的次数。这就是我从数组传递ArrayList到函数创建新数组列表并删除已经匹配的元素的原因。我已经将2个功能改为一个，因为我认为这是编写代码的更好方法。

Answer 2

完全不同的方法：

void disjoint(String s1, String s2) {
    // Convert two strings to character lists
    List<Character> chars1 = s1.chars()
        .mapToObj(t -> (char) t)
        .collect(Collectors.toList());
    List<Character> chars2 = s2.chars()
        .mapToObj(t -> (char) t)
        .collect(Collectors.toList());

    // Make a list of elements we are going to remove from chars1
    List<Character> removedElements = new ArrayList<>();
    chars1.forEach(t -> {
        // Try to remove the current character from chars2, and if it has
        // successfully been removed, add it to our removedElements list
        if (chars2.remove(t)) {
            removedElements.add(t);
        }
    });
    // At last remove each removed element from chars2 also from chars1
    removedElements.forEach(chars1::remove);

    System.out.println(chars1);
    System.out.println(chars2);

}

将数组与重复元素进行比较，并将不存在的元素添加到相应的数组中，包括重复元素

2 个答案: