如何找到最近的回文数？

Question

好吧，我试着在 java 中编写一个程序来查找回文，有人让我在其中添加一个扭曲，扭曲是这样的，我们必须找出数字是否是回文，如果不是那么我们必须找到最近的回文数。例如，如果输入的数字是 37，那么在它附近有两个可以是回文的数字，一个是 33，另一个是 44，但较近的是 33，所以这里的答案是 33。我尝试了很多但没有成功:(

import java.io.*;  
public class NearestPalindrome {  
static int i,j,n,m,f,r=0,g,p1,p2,c1=0,c2=0,a;  
static boolean k;  
public static void main (String []agrs)throws IOException  
{
   BufferedReader A = new BufferedReader(new InputStreamReader(System.in));  
   System.out.println("This Program Is Created For Checking Whether The Number Entered Is Palindrome, If Not Find The Nearest Palindrome..");  
   System.out.println("Please Enter Number:-");  
   n=Integer.parseInt(A.readLine());
   f=n;  
   NearestPalindrome.Palindrome(f);  
   if(r==f)
   {System.out.println("This Is A Palindrome Number:-"+r);}
   else
   {
       for(i=n;((c1==1)&&(c2==1));i--)
       {
           NearestPalindrome.Palindrome(i);  
           NearestPalindrome.Palindrome(j);  
           if(k=true)  
           { if(r==i)  
           {   p1=i;  
           c1++;  
           }  
           if(r==j)  
               {p2=j;  
               c2++;  
               }  
               }  
           }  
       if((n-p1)>(p2-n))    
       {
        System.out.println("The Nearest Palindrome Is:-"+p2);  
       }  
       else  
        System.out.println("The Nearest Palindrome Is:-"+p1);  

       }  

   }  
public static boolean Palindrome (int x)
{


    for(i=0;x>0;i++)
       {
           m=x%10;
           r=r*10+m;
           x=x/10;   
       }  
    if(n==r)
    {k=true;}
    else
        k=false;
return k;
}

}

Answer 1

这会让您朝着正确的方向前进

public class MyClass {

    public int findNearestPalindrome(int numberEntered) {
        int lowerPalidrome = getLowerPalindrome(numberEntered);
        int higherPalindrome = getHigherPalindrome(numberEntered);
        if (Math.abs(numberEntered - lowerPalindrome) < Math.abs(numberEntered - higherPalindrome) {
            return lowerPalindrome;
        } else {
            return higherPalindrom;
        }
    }

    public int getLowerPalindrome(int baseNumber) {
        for (int i = baseNumber; i > 0; i--) {
            //do logic to find palindrome and return it
        }
    }

    public int getHigherPalindrome(int baseNumber) {
        for (int i = baseNumber; i < LARGE_NUMBER; i++) {
            //do logic to find palindrome and return it
        }
    }
}

请记住，您可以通过多种方式做到这一点。例如，您可以使用 while 循环，而不是 for 循环。希望对您有所帮助！

编辑：如果你真的想压缩代码，我建议你创建一个找到回文的方法，并在两种方法中使用它。

Answer 2

可以更高效地完成。

假设，我们有 N 个数字。让我们找到长度为 N / 2 的前缀。之后，您可以使用 2 个最近的前缀构造 3 个可能的答案，并找到最近的回文（您应该在这里考虑极端情况）。

示例：

号码 123456789 有 1234 个长度为 9/2 = 4 的前缀。最近的前缀：1233、1235.

然后我们只构建 3 个候选项：123353321、123454321 和 123555321

abs(123456789 - 123353321) = 103468
abs(123456789 - 123454321) = 2468
abs(123456789 - 123555321) = 98532

所以最近的回文将是123454321。

UPD：在这种情况下，您也必须尝试所有可能的中间数字。

Answer 3

package com.amit.practice;

public class Test {

    public static void main(String[] args) {
        Test test= new Test();
        int polyNum=test.polydrum(813123);
        System.out.println(polyNum);
    }

    public int polydrum(int number){
        int finalNum=0;
        System.out.println("The given number is :"+number+" ");
        do {
            String revString = "";
            String numString=""+number;
            int numLength=numString.length();

            for(int i=numLength-1;i>=0;i--){
                revString += numString.charAt(i);
                if(numString.equals(revString)){
                    System.out.println("The next polydrum number is :"+numString);
                    finalNum=Integer.parseInt(revString);
                    return finalNum;
                }
            }
            number=number+1;
        } while(finalNum!=number);

        return finalNum;
    }
}

Answer 4

这段代码是使用nearestPalindrome(int number)方法找到最近的回文数，其他方法使用nextPalindrome(int number)找到下一个回文数。

package com.amit.palindrome;

public class Palindrome {

    public static void main(String[] args) {
        Palindrome test= new Palindrome();
        int number=135625978;
        System.out.println("Given Number is :"+number);

        int nearestPolyNum=test.nearestPalindrome(number);
        int nextPolyNum=test.nextPalindrome(number);
        System.out.println("Nearest Palindrome Number is :"+nearestPolyNum);
        System.out.println("Next Palindrome Number is :"+nextPolyNum);
    }

    public int nextPalindrome(int number){
        int nextPoly=0;
        if(number>9){
            do {
                String revString = "";
                String numString=""+number;
                int numLength=numString.length();

                for(int i=numLength-1;i>=0;i--){
                    revString += numString.charAt(i);
                    if(numString.equals(revString)){
                        nextPoly=Integer.parseInt(revString);
                        return nextPoly;
                    }
                }
                number=number+1;
            } while(nextPoly!=number);
            return nextPoly;
        }
       return number;
    }



    public int nearestPalindrome(int number){

        if(number>9){
                String revString = "";
                String numString=""+number;
                int nearestPoly=0;
                int numLength=numString.length();           

                    String anotherHalfNumber=null;
                    int halfNum = numLength/2;
                    String firstHalfNumber=numString.substring(0, halfNum);
                    if(numLength%2==0){
                        anotherHalfNumber=numString.substring(halfNum);
                    }else{
                        anotherHalfNumber=numString.substring(halfNum+1);
                    }               
                    for(int i=firstHalfNumber.length()-1;i>=0;i--){
                        revString += firstHalfNumber.charAt(i);                

                    }
                    int intAnotherHalfNumber=Integer.parseInt(anotherHalfNumber);
                    int reverseFirstHalfNumber=Integer.parseInt(revString);
                    int adjustedNumber = intAnotherHalfNumber-reverseFirstHalfNumber;
                    nearestPoly = number-adjustedNumber;
                    return nearestPoly;
        }
         return number;
    }
}

Answer 5

    import java.util.*;
    import java.lang.*;
    import java.io.*;
    class GFG
     {  public static int findcount(int n)
     {
         int count=0;
         while(n!=0)
         {
             n=n/10;
             count++;
         }
         return count;
     }
public static int findpalindrome(int arr)
 {
     int rem=0,count=0;
    while(arr>0)
    {
        rem=arr%10;
        count=(count*10)+rem;
        arr=arr/10;
    }
     return count;
 }
public static void main (String[] args) 
 {
 //code
 Scanner sc=new Scanner(System.in);
 int test=sc.nextInt();

 while(test-->0)
 { 
    int n=sc.nextInt();
     int res=0,flag=0,flag11=0,j=n+1,i=n-1;
     int aa=0,bb=0;
     if(findcount(n)==1 || (n==findpalindrome(n)) )
        System.out.println(n);
     else
     { 

         for(i=n-1;i>0;i--)
         {

          if(i==findpalindrome(i))
          {    
              aa=i;
              break;
          }  

         }
         while(flag!=1)
         { 
          if(j==findpalindrome(j))
          {    
              bb=j;
              flag=1;
              break;
          }
         j++;  
         } 
       int dif1=Math.abs(n-aa);
       int dif2=Math.abs(n-bb);
       if((dif1==dif2) || (dif1<dif2))
        System.out.println(aa);
       else
       System.out.println(bb);
     }

 }
 }

}

Answer 6

使用带有数组和 StringBuilder 的基本 Java 的代码

    package newPackage;

public class AmazonQn6FindNearestPalindromeNumber {
    static int before;
    static int after;
    public static void main(String[] args) {
        System.out.println();

        int num=1221;
        AmazonQn6FindNearestPalindromeNumber n1=new AmazonQn6FindNearestPalindromeNumber();

        before = num-1;
        after = num+1;


        int[] checkPalindrome= new int[2];

        do {
            checkPalindrome = n1.checkPalindrome(before, after);        
            if((checkPalindrome[0] >0)||(checkPalindrome[1] >0)) {
            System.out.println("Numbers are");
            for(int n:checkPalindrome) {
                System.out.println(n);
            }


            if(checkPalindrome[0]!=0)
                System.out.println("Nearest Palindrome is "+checkPalindrome[0]);

            if(checkPalindrome[1]!=0)
                System.out.println("Nearest Palindrome is "+checkPalindrome[1]);

            break;
            }
        //  checkPalindrome = n1.checkPalindrome(before, after);

        }while((checkPalindrome[0] ==0)||(checkPalindrome[1] ==0));

    }

    public int[] checkPalindrome(Integer n, Integer m) {
        before = before-1;
        after = after+1;
        int[] array = new int[2];

        String stringn = n.toString();
        StringBuilder string1 = new StringBuilder(stringn);
        System.out.println(string1);
        StringBuilder string2 = string1.reverse();
        System.out.println(string2);
        if(stringn.equals(string2.toString()))
        {   
            array[0]=n;
            System.out.println("Element in n" +array[0]);

        }

        String stringm = m.toString();
        StringBuilder string3 = new StringBuilder(stringm);
        System.out.println(string3);
        StringBuilder string4 = string3.reverse();
        System.out.println(string4);

        if(stringm.equals(string4.toString()))
        {
            array[1]=m;
            System.out.println("Element in m" +array[1]);

        }

        return array;

    }

}

Answer 7

查找最近回文的最简单代码。例如，如果输入是 37。最接近 37 的回文是 33 和 44。但分别是 4 和 7。因此输出为 33。如果两种情况下的差异相等，它将打印两个数字。

package demo;

import java.util.Scanner;

import javax.swing.text.html.FormSubmitEvent;

public class Increment {


    public static boolean Plaindrome(int n) {
          int r,sum=0,temp;

          temp=n;    
          while(n>0){    
           r=n%10;  
           sum=(sum*10)+r;    
           n=n/10;    
          }    
          if(temp==sum)    
              return true;
          else    
           return false; 

        }  
    public static int  greaterpalindrome(int k) {
        while(!Plaindrome(k)) {

             k++;

         }
        return k;
    }
    public static int  smallerpalindrome(int s) {
        while(!Plaindrome(s)) {

            s--;

         }
        return s;}

    public static void main(String[] args) {
         Scanner scan = new Scanner(System.in);
         int value = scan.nextInt();

         int greater = greaterpalindrome(value+1);

         int smaller = smallerpalindrome(value-1);

         if (Math.abs(value-greater)==(value-smaller)) {
                System.out.println(greater+" "+smaller);
            }
            else if (Math.abs(value-greater)>(value-smaller)) {
                System.out.println(smaller);
            }
            else {
                System.out.println(greater);
            }

        } }

Answer 8

试试下面的代码片段：

public class Sample{
    public static void main(String[] args) throws InterruptedException {
        Scanner scan = new Scanner(System.in);
        
        System.out.print("Enter val: ");
        
        int entry = scan.nextInt();
        int entryN = entry;
        int entryP = entry;

        while(true) {
            if (isPolindrom(entry)) {
                System.out.println(entry);
                break;
            } else if (isPolindrom(--entryN)) {
                System.out.println(entryN);
                break;
            } else if (isPolindrom(++entryP)){
                System.out.println(entryP);
                break;
            }
        }   
        
        System.out.println("Best hakiki mürşit is a science :)");
        
    }

    private static boolean isPolindrom(int a) {
        int val1 = a;
        int last = 0;
        int size = String.valueOf(val1).length();
        
        for(int i = size; i > 0 ; i-- ) {
            
            last += (val1 % 10) * (int) Math.pow(10, (i-1));
            val1 = (val1 - (val1 % 10)) /10;
        }
        if (a == last) {
            return true;
        } else {
            return false;
        }       
        
    }
}

如何找到最近的回文数？

How To Find The Nearest Palindrome Number?

java

algorithm

palindrome