Question

我试图为像BigInteger这样的类实现一个add函数，其中输入是一个疯狂的长数字。但是，当我从字节数组转换为字符串时，屏幕上没有任何内容正在打印。但是我在main函数中注释掉的示例代码可以正常工作。有谁知道为什么？

以下是coderpad上的输出 resut字节数组：[7,9,1,0]
100 + 9 =

import java.io.*;
import java.util.*;
import java.nio.charset.Charset;

class Solution {
  private static Charset charset = Charset.forName("UTF-8");

  public static class BigInteger {
    public byte[] digits;

    public BigInteger(String input) {
      digits = input.getBytes(charset);
    }
    public BigInteger(byte[] digits) {
        this.digits = digits;
//         System.out.println(Arrays.toString(digits));
    }

   public String add(BigInteger other) {
        byte[] digits1, digits2;
        if (this.digits.length >= other.digits.length) {
            digits1 = this.digits;
            digits2 = other.digits;
        } else {
            digits1 = other.digits;
            digits2 = this.digits;
        }
        int remainder = 0;
        int s1 = digits1.length - 1;
        int s2 = digits2.length - 1;
        byte[] resultDigits = new byte[s1 + 2];

        for (int i = s1; i >= 0; i--) {
            int digit1 = digits1[i];
            int digit2 = i > s2 ? 0 : digits2[i];
            int r = digit1 + digit2 + remainder;
            remainder = r > 9 ? 1 : 0;
            resultDigits[i] = (byte) (r % 10);
//             System.out.println(resultDigits[i]);
        }
        resultDigits[s1] = (byte) remainder;

//         System.out.println("resut byte array: " + Arrays.toString(resultDigits));
     String result = new String(resultDigits, charset);
      return result;
    }

  }
  public static void main(String[] args) {
    BigInteger a = new BigInteger("100");
    BigInteger b = new BigInteger("9");
    System.out.println("100 + 9 = " + a.add(b));

//     String abc="123";

//     byte[] ba = abc.getBytes();
//     System.out.println(Arrays.toString(ba));
//     System.out.println(new String(ba, Charset.forName("UTF-8")));
  }
}

---编辑 - add（）v2

public String add(BigInteger other) {
        byte[] digits1, digits2;
        if (this.digits.length >= other.digits.length) {
            digits1 = this.digits;
            digits2 = other.digits;
        } else {
            digits1 = other.digits;
            digits2 = this.digits;
        }
        int remainder = 0;
        int s1 = digits1.length - 1;
        int s2 = digits2.length - 1;
        byte[] resultDigits = new byte[s1 + 2];

        for (int i = s1; i >= 0; i--) {
            int digit1 = digits1[s1--];
            int digit2 = s2 >= 0 ? digits2[s2--] : 0;
            int sum = digit1 + digit2 + remainder;
            remainder = sum / 10;
            resultDigits[i] = (byte) (sum % 10);
//             System.out.println(resultDigits[i]);
        }
        resultDigits[0] = (byte) remainder; // probably dont need this

        System.out.println("resut byte array: " + Arrays.toString(resultDigits));
     String result = new String(resultDigits, charset);
     System.out.println("result as string: " + result);
      return result;
    }

Answer 1

以下是更正：

public String add(BigInteger other) {
            byte[] digits1, digits2;
            if (this.digits.length >= other.digits.length) {
                digits1 = this.digits;
                digits2 = other.digits; 
            } else {
                digits1 = other.digits;
                digits2 = this.digits;
            }
            int remainder = 0;
            int s1 = digits1.length - 1; 
            int s2 = digits2.length - 1; 
            byte[] resultDigits = new byte[s1 + 2];

            for (int i = s1; i >= 0; i--) {
                int diff = digits1.length - digits2.length;
                int digit1 = digits1[i] - 48;
                int digit2 = i < diff ? 0 : digits2[i - diff] - 48;
                int r = digit1 + digit2 + remainder; 
                remainder = r > 9 ? 1 : 0;
                resultDigits[i+1] = (byte)(r % 10 + 48);
            }
            resultDigits[0] += (byte) remainder + 48;
            String result = resultDigits[0] == 48 ? new String(resultDigits).substring(1) : new String(resultDigits);
            return result;
        }

基本上，错误是：

你知道自己正在输入字符
您的小操作数的位置也遇到了麻烦。
最后，你在最后添加的剩余部分被添加到一个不好的位置。您应该始终将其添加到第一个索引。

输出：

0 + 45 = 45
19 + 95 = 114
38 + 20 = 58
57 + 92 = 149
76 + 67 = 143
95 + 54 = 149
114 + 90 = 204
133 + 48 = 181
152 + 6 = 158
171 + 18 = 189
190 + 22 = 212
209 + 13 = 222
228 + 32 = 260
247 + 24 = 271

java biginteger使用byte []

1 个答案: