如何将具有不同位数的整数的数组表示形式相加？

Question

How would I add together two integers with different number of digits, using an array, without causing an out of bounds exception?

For example: 500 + 99 each digit is an element of the array

This is how I'm doing it right now:

 int aIILength = infiniteInteger.length-1;
 int bIILength = anInfiniteInteger.infiniteInteger.length-1;
 for(int f = aIILength; f >0; f--){

        int aTempNum = infiniteInteger[f];


        int bTempNum = anInfiniteInteger.infiniteInteger[f];

        result = aTempNum + bTempNum;

        //array representation of sum
        tempArray[f] = result;

    }

Answer 1

Let the counter in the loop go from 1 and up, and use it to access the digits from the end of each array.

You need a carry to hold the overflow of adding each set of digits.

Loop until you run out of digits in both arrays, and carry is zero.

Check the range when you access the digits from the arrays, and use zero when you run out of digits.

int aIILength = infiniteInteger.length;
int bIILength = anInfiniteInteger.infiniteInteger.length;
int carry = 0;
for(int f = 1; f <= aIILength || f <= bIILength || carry > 0; f++){
  int aTempNum;
  if (f <= aIILength) {
    aTempNum = infiniteInteger[aIILength - f];
  } else {
    aTempNum = 0;
  }
  int bTempNum;
  if (f <= bIILength) {
    bTempNum = anInfiniteInteger.infiniteInteger[bIILength - f];
  } else {
    bTempNum = 0;
  }
  result = aTempNum + bTempNum + carry;
  tempArray[tempArray.length - f] = result % 10;
  carry = result / 10;
}

Note: Make tempArray longer than both the operand arrays, so that it has place for a potential carry to the next digit.