Java 在通过删除字符而不是删除次数使两个字符串变位后打印它们？

>在这里，我可以打印字符的重复次数，但我不能 能够得到如何比较和删除。这里我为两个字符串和键取了两个映射作为字符，值作为重复

public类noofdeletionTosthTringAngram{
@抑制警告（“资源”）
公共静态void main（字符串[]args）{
System.out.println（“字符串”）；
扫描仪sc=新的扫描仪（System.in）；
System.out.println（“输入字符串1”）；
字符串s1=sc.next（）；
System.out.println（“输入字符串2”）；
字符串s2=sc.next（）；
Map m1s1=新的HashMap（）；
Map m2s2=新的HashMap（）；
对于（int i=0；i

Samlie输入： s1=abc s2=cde 在这里，我们必须通过从两个字符串中删除字符来进行字谜分析 输出： s1=c s2=c

/*任何包；//不要放置包名*/
    /* package whatever; // don't place package name! */

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



    class Ideone
    {

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

        System.out.println("string");
        Scanner sc = new Scanner(System.in);
        System.out.println("enter string 1");
        String s1 = sc.next();
        System.out.println("enter string 2");
        String s2 = sc.next();
        Map<Character, Integer> m1s1 = new HashMap<>();
        Map<Character, Integer> m2s2 = new HashMap<>();

        for (int i = 0; i < s1.length(); i++) {

            if (m1s1.containsKey(s1.charAt(i))) {
                m1s1.put((Character) s1.charAt(i), m1s1.get((Character) 
        s1.charAt(i)) + 1);
            } else {

                m1s1.put((Character) s1.charAt(i), 1);

            }
        }
        for (int i = 0; i < s2.length(); i++) {

            if (m2s2.containsKey(s2.charAt(i))) {
                m2s2.put((Character) s2.charAt(i), m2s2.get((Character) 
         s2.charAt(i)) + 1);
            } else {
                m2s2.put((Character) s2.charAt(i), 1);

            }
        }
        System.out.println("m1s1....." + m1s1);
        System.out.println("m221...." + m2s2);

// ADDED MY CODE FROM HERE
// TAKEN STRING 1 AND MAP OF STRING 2, CHECK CHARACTER BY CHARACTER OF STRING 1 IN MAP //OF STRING 2, IF THE CHARACTER EXISTS PRINT IT & DECREASE ITS VALUE IN MAP BY 1 , IF //AFTER DECREASING THE VALUE BECOMES 0, REMOVE IT FROM MAP, IF THE CHARACTER OF //STRING1 DOES NOT EXIST IN MAP2 DON'T DO ANYTHING, FOLLOW ABOVE RULES FOR EVERY //CHARACTER OF STRING 1

// NOW TAKE STRING2 AND MAP OF STRING1 AND FOLLOW THE EXACT PROCEDURE ABOVE

        int i,x;
        char c;

        for(i=0;i<s1.length();i++)
        {
            c=s1.charAt(i);
            if(m2s2.containsKey(c))
            {
                System.out.print(c);
                x=m2s2.get((Character)c);
                x=x-1;
                if(x==0)
                    m2s2.remove(new Character(c));
                else
                    m2s2.put((Character) c,x);
            }
        }
                System.out.println();

            for(i=0;i<s2.length();i++)
        {
            c=s2.charAt(i);
            if(m1s1.containsKey(c))
            {
                System.out.print(c);
                x=m1s1.get((Character)c);
                x=x-1;
                if(x==0)
                    m1s1.remove(new Character(c));
                else
                    m1s1.put((Character) c,x);
            }
        }


        }
    }

导入java.util.*；
导入java.lang.*；
导入java.io.*；
表意文字
{
公共静态void main（字符串[]args）引发java.lang.Exception
{
System.out.println（“字符串”）；
扫描仪sc=新的扫描仪（System.in）；
System.out.println（“输入字符串1”）；
字符串s1=sc.next（）；
System.out.println（“输入字符串2”）；
字符串s2=sc.next（）；
Map m1s1=新的HashMap（）；
Map m2s2=新的HashMap（）；
对于（int i=0；i对于（i=0；我不知道你们为什么投反对票。我也通过发布代码来询问我的疑问。你能在你的问题中添加示例输入和输出吗？不清楚你在问什么。@NicholasK示例输入和输出添加了什么问题？你被困在哪里了？我不知道如何迭代删除字符以使anagramThanks变得非常@anupam、 .非常有用
    /* package whatever; // don't place package name! */

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



    class Ideone
    {

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

        System.out.println("string");
        Scanner sc = new Scanner(System.in);
        System.out.println("enter string 1");
        String s1 = sc.next();
        System.out.println("enter string 2");
        String s2 = sc.next();
        Map<Character, Integer> m1s1 = new HashMap<>();
        Map<Character, Integer> m2s2 = new HashMap<>();

        for (int i = 0; i < s1.length(); i++) {

            if (m1s1.containsKey(s1.charAt(i))) {
                m1s1.put((Character) s1.charAt(i), m1s1.get((Character) 
        s1.charAt(i)) + 1);
            } else {

                m1s1.put((Character) s1.charAt(i), 1);

            }
        }
        for (int i = 0; i < s2.length(); i++) {

            if (m2s2.containsKey(s2.charAt(i))) {
                m2s2.put((Character) s2.charAt(i), m2s2.get((Character) 
         s2.charAt(i)) + 1);
            } else {
                m2s2.put((Character) s2.charAt(i), 1);

            }
        }
        System.out.println("m1s1....." + m1s1);
        System.out.println("m221...." + m2s2);

// ADDED MY CODE FROM HERE
// TAKEN STRING 1 AND MAP OF STRING 2, CHECK CHARACTER BY CHARACTER OF STRING 1 IN MAP //OF STRING 2, IF THE CHARACTER EXISTS PRINT IT & DECREASE ITS VALUE IN MAP BY 1 , IF //AFTER DECREASING THE VALUE BECOMES 0, REMOVE IT FROM MAP, IF THE CHARACTER OF //STRING1 DOES NOT EXIST IN MAP2 DON'T DO ANYTHING, FOLLOW ABOVE RULES FOR EVERY //CHARACTER OF STRING 1

// NOW TAKE STRING2 AND MAP OF STRING1 AND FOLLOW THE EXACT PROCEDURE ABOVE

        int i,x;
        char c;

        for(i=0;i<s1.length();i++)
        {
            c=s1.charAt(i);
            if(m2s2.containsKey(c))
            {
                System.out.print(c);
                x=m2s2.get((Character)c);
                x=x-1;
                if(x==0)
                    m2s2.remove(new Character(c));
                else
                    m2s2.put((Character) c,x);
            }
        }
                System.out.println();

            for(i=0;i<s2.length();i++)
        {
            c=s2.charAt(i);
            if(m1s1.containsKey(c))
            {
                System.out.print(c);
                x=m1s1.get((Character)c);
                x=x-1;
                if(x==0)
                    m1s1.remove(new Character(c));
                else
                    m1s1.put((Character) c,x);
            }
        }


        }
    }