逐个元素地在数组中添加数字

Question

The goal is to add two numbers together that are stored in arrays-element by element. The numbers do not necessarily need to be of equal length. I am having trouble accounting for the possibility of a carry over.

If the number is 1101 it would be represented : [1,0,1,1]- The least significant bit is at position 0. I am doing the addition without converting it to an integer.

I am making a separate method to calculate the sum of binary numbers but I just want to understand how to go about this using the same logic.

Ex: 349+999 or they could even be binary numbers as well such as 1010101+11

Any suggestions?

    int carry=0;
    int first= A.length;
    int second=B.length;
    int [] sum = new int [(Math.max(first, second))];
    if(first > second || first==second)
    {
        for(int i =0; i <A.length;i++)
        {
            for(int j =0; j <B.length;j++)
            {
                sum[i]= (A[i]+B[j]);
            }
        }
        return sum; 
    }
    else 
    {
        for(int i =0; i <B.length;i++)
        {
            for(int j =0; j <A.length;j++)
            {
                sum[i]= (A[i]+B[j]);
            }
        }
        return sum;
    }

For the binary addition:

   byte carry=0;
    int first= A.length;
    int second=B.length;
    byte [] sum = new byte [Math.max(first, second)+1];
    if(first > second || first==second)
    {
        for(int i =0; i < A.length && i!= B.length ;i++)
        {
            sum[i]= (byte) (A[i] + B[i] + carry);
            if(sum[i]>1) {
                sum[i] = (byte) (sum[i] -1);
                carry = 1;
            }
            else
                carry = 0;
        }
        for(int i = B.length; i < A.length; i++) {
           sum[i] = (byte) (A[i] + carry);
           if(sum[i]>1) {
                sum[i] = (byte) (sum[i] -1);
                carry = 1;
            }
            else
                carry = 0;
       }
        sum[A.length] = carry; //Assigning msb as carry
        return sum; 
    }
    else 
    { 
        for(int i =0; i < B.length && i!= A.length ;i++) {
            sum[i]= (byte) (A[i] + B[i] + carry);
            if(sum[i]>1) {
                sum[i] = (byte) (sum[i] -1);
                carry = 1;
            }
            else
                carry = 0;
        }
        for(int i = A.length; i < B.length; i++) {
           sum[i] = (byte) (B[i] + carry);
           if(sum[i]>1) {
                sum[i] = (byte) (sum[i] -1);
                carry = 1;
            }
            else
                carry = 0;
       }
        sum[B.length] = carry;//Assigning msb as carry
        return sum; 
    }

Answer 1

There is no need to treat binary and decimal differently. This handles any base, from binary to base36, and extremely large values -- far beyond mere int's and long's!

Digits need to be added from the least significant first. Placing the least significant digit first makes the code simpler, which is why most CPUs are Little-Endian.

Note: Save the code as "digits.java" -- digits is the main class. I put Adder first for readability.

Output:

NOTE: Values are Little-Endian! (right-to-left)
base1: 0(0) + 00(0) = 000(0)
base2: 01(2) + 1(1) = 110(3)
base2: 11(3) + 01(2) = 101(5)
base2: 11(3) + 011(6) = 1001(9)
base16: 0A(160) + 16(97) = 101(257)
base32: 0R(864) + 15(161) = 101(1025)

Source code: digits.java:

class Adder {
  private int base;
  private int[] a;
  private int[] b;
  private int[] sum;

  public String add() {
    int digitCt= a.length;
    if(b.length>digitCt)
      digitCt= b.length;        //max(a,b)
    digitCt+= 1;                //Account for possible carry
    sum= new int[digitCt];      //Allocate space
    int digit= 0;               //Start with no carry
    //Add each digit...
    for(int nDigit=0;nDigit<digitCt;nDigit++) {
      //digit already contains the carry value...
      if(nDigit<a.length)
        digit+= a[nDigit];
      if(nDigit<b.length)
        digit+= b[nDigit];
      sum[nDigit]= digit % base;//Write LSB of sum
      digit= digit/base;        //digit becomes carry
    }
    return(arrayToText(sum));
  }

  public Adder(int _base) {
    if(_base<1) {
      base= 1;
    } else if(_base>36) {
      base=36;
    } else {
      base= _base;
    }
    a= new int[0];
    b= new int[0];
  }

  public void loadA(String textA) {
    a= textToArray(textA);
  }

  public void loadB(String textB) {
    b= textToArray(textB);
  }

  private int charToDigit(int digit) {
    if(digit>='0' && digit<='9') {
      digit= digit-'0';
    } else if(digit>='A' && digit<='Z') {
      digit= (digit-'A')+10;
    } else if(digit>='a' && digit<='z') {
      digit= (digit-'a')+10;
    } else {
      digit= 0;
    }
    if(digit>=base)
      digit= 0;
    return(digit);
  }

  private char digitToChar(int digit) {
    if(digit<10) {
      digit= '0'+digit;
    } else {
      digit= 'A'+(digit-10);
    }
    return((char)digit);
  }

  private int[] textToArray(String text) {
    int digitCt= text.length();
    int[] digits= new int[digitCt];
    for(int nDigit=0;nDigit<digitCt;nDigit++) {
      digits[nDigit]= charToDigit(text.charAt(nDigit));
    }
    return(digits);
  }

  private String arrayToText(int[] a) {
    int digitCt= a.length;
    StringBuilder text= new StringBuilder();
    for(int nDigit=0;nDigit<digitCt;nDigit++) {
      text.append(digitToChar(a[nDigit]));
    }
    return(text.toString());
  }

  public long textToInt(String a) {
    long value= 0;
    long power= 1;
    for(int nDigit=0;nDigit<a.length();nDigit++) {
      int digit= charToDigit(a.charAt(nDigit));
      value+= digit*power;
      power= power*base;
    }
    return(value);
  }
}

public class digits {

  public static void main(String args[]) {
    System.out.println("NOTE: Values are Little-Endian! (right-to-left)");
    System.out.println(test(1,"0","00"));
    System.out.println(test(2,"01","1"));
    System.out.println(test(2,"11","01"));
    System.out.println(test(2,"11","011"));
    System.out.println(test(16,"0A","16"));
    System.out.println(test(32,"0R","15"));
  }

  public static String test(int base, String textA, String textB) {
    Adder adder= new Adder(base);
    adder.loadA(textA);
    adder.loadB(textB);
    String sum= adder.add();
    String result= String.format(
      "base%d: %s(%d) + %s(%d) = %s(%d)",
       base,
       textA,adder.textToInt(textA),
       textB,adder.textToInt(textB),
       sum,adder.textToInt(sum)
    );
    return(result);
  }

}

Answer 2

First off, adding binary numbers will be different then ints. For ints you could do something like

int first = A.length;
int second = B.length;

int firstSum = 0;
for (int i = 0; i < first; i++){
    firstSum += A[i] * (10 ^ i);
}

int secondSum = 0;
for (int j = 0; j < second; j++){
    secondSum += B[j] * (10 ^ j);
}

int totalSum = firstSum + secondSum;

Answer 3

It should be represented like this in below as we have to add only the values at same position of both the arrays. Also, in your first if condition, it should be || instead of && . This algorithm should perfectly work. Let me know if there are any complications.

int carry=0;
int first= A.length;
int second=B.length;
int [] sum = new int [Math.max(first, second)+1];
if(first > second || first==second)
{
    for(int i =0; i < A.length && i!= B.length ;i++)
    {
        sum[i]= A[i] + B[i] + carry;
        if(sum[i]>9) {
            sum[i] = sum[i] -9;
            carry = 1;
        }
        else
            carry = 0;
    }
    for(int i = B.length; i < A.length; i++) {
       sum[i] = A[i] + carry;
       if(sum[i]>9) {
            sum[i] = sum[i] -9;
            carry = 1;
        }
        else
            carry = 0;
   }
    sum[A.length] = carry; //Assigning msb as carry
    return sum; 
}
else 
{ 
    for(int i =0; i < B.length && i!= A.length ;i++) {
        sum[i]= A[i] + B[i] + carry;
        if(sum[i]>9) {
            sum[i] = sum[i] -9;
            carry = 1;
        }
        else
            carry = 0;
    }
    for(int i = A.length; i < B.length; i++) {
       sum[i] = B[i] + carry;
       if(sum[i]>9) {
            sum[i] = sum[i] -9;
            carry = 1;
        }
        else
            carry = 0;
   }
    sum[B.length] = carry //Assigning msb as carry
    return sum; 
}