Question

所以我正在创建一个基于this项目制作回文数字的程序。我的程序适用于较小的数字，但整数只能计算少量数字，因此我将必要的整数更改为BigInteger（）。在做完这个之后，我遇到了一些我不太确定的问题。有人会对如何使这项工作有任何建议吗？

public class Main {
public static final boolean DEBUG = false;

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

    //gets the number from the console
    int number = input.nextInt();
    if(DEBUG) System.out.println("Got your number!");
    if(DEBUG) System.out.println("Bout to makePalendrome!");

    makePalendrome(number);   
}


public static boolean isPalendrome(BigInteger number){
    String numberString = number.toString();

    for(int i = 0; i < numberString.length(); i++){
        if(numberString.charAt(i) != numberString.charAt(numberString.length() - 1 - i)) return false;   
    }       
    return true;
}


public static void makePalendrome(int input){
    int steps = 0;
    BigInteger number = new BigInteger((input + "")); 
    if(isPalendrome(number)) printResult(input, steps, number);

    while(!isPalendrome(number)){
        String numberString = number.toString();
        String reversed = "";
        for(int i = 0; i < numberString.length(); i++){
            reversed += numberString.charAt(numberString.length() - 1 - i);
        }
        BigInteger numReversed = new BigInteger(reversed);
        number.add(numReversed);
        steps++;
    }

    printResult(input, steps, number);
}

public static void printResult(int number, int steps, BigInteger palendrome){
        System.out.printf("The number %d becomes palendromic after %d steps, and becomes the number: %d%n", number, steps, palendrome);
        System.exit(0);
    }

}

Answer 1

您的代码进入无限循环while(!isPalendrome(number))因为，

  number.add(numReversed);

这不会改变number的值。您需要将其分配回来。

number= number.add(numReversed);

切换到BigInteger并且progam不再运行

1 个答案: