/* Online Java - IDE, Code Editor, Compiler Online Java is a quick and easy tool that helps you to build, compile, test your programs online. */ //I have declared all variables class level and static to ensure //that the recursive method is kept tidy //the question also raises performance question //to ensure I can test the execution time, I will add stopwatch.. //For moment, I am using long variable to ensure I can extend //the String to 9,223,372,036,854,775,807 public class Main { static int lengthFirstNumber; static int lengthSecondNumber; static String lastDigitFirstNumber; static String lastDigitSecondNumber; static int firstDigitTotalDigits; static int lastDigitTotalDigits; static int total; static String grandTotal=""; static boolean remainder=false; static String remainingPortion; static int remainingPortionInteger; static String remainingPortionString; public static void main(String[] args) { System.out.println("Welcome to Online IDE!! Happy Coding :)"); System.out.println("Addition program, ensure String has maximum characters: " + "2,147,483,648"); //Note it is important to have the numbers structured like this also, //otherwise the code will fail. //String number2 = "59"; //String number1 = "652"; //String number2 = " 544"; //String number1 = "4666"; String number2 = "84636"; String number1 = " 324"; //For an unknown reason, the addition method is unable to return the String, it just shows blank value //Bizarre is that it can see the variable value on screen before I try to return it. //So I have just used System.out.println(GRANDTOTAL). System.out.println(number1 + "+" + number2 + "= " + addition(number1, number2)); } public static String addition(String number1, String number2) { //it is still working constraints of the code //the Strings are note being converted, it is only the length of the Strings. lengthFirstNumber=(number1.length()); lengthSecondNumber=(number2.length()); /* //at this point it knowws that its on the most extreme left hand side of them... //looking at example 4666 // + 544 //it is just not a case of dropping the number 4 down.... into the grandTotal //this would be ok if (6+5 was not >=10) //there would be last addition undertaken //at this point its expected to perform 4 + //another example is just a case of dropping the remainder across //for instance performing 55 + 55 ------- 110 ------- //Another example is dealing with larger differences: 274666 + 544 //it would need to carry 1, add it to 4 //it would then have to truncate number1 again and drop 2 and 7 */ //if both lengths are 0, it means both had same lengths. if (lengthFirstNumber==0 && lengthSecondNumber==0) { //this is a slightly ambiguous name, but it identifies if //total from digit in number1 number2 is one or two digits long //firstDigitTotalDigits = (int)total/10; //if there is no frontal number (i.e) total less than 10, expect 0 in firstDigitTotalDigits //if total is between 10 - 18(9+9), expect firstDigitTotalDigits=1 if (firstDigitTotalDigits==0) { System.out.println("GRANDTOTAL: " + grandTotal); return grandTotal; } else { //it would append a 1 to the left of existing most significant digit grandTotal = "1" + grandTotal; System.out.println("GRANDTOTAL: " + grandTotal); return grandTotal; } } //unfortunately variable names are extremely confusing //need to remember it takes values from here onwards with one argument //if number1 was 9123 lengthFirstNumber=4 //there is a digit at index 3, so variable is reduced by 1 lastDigitFirstNumber=number1.substring(lengthFirstNumber-1); lastDigitSecondNumber=number2.substring(lengthSecondNumber-1); //need something here since it was giving numberformat exception. //most likely since if no value in Strings (lastDigitFirstNumber //or lastDigitSecondNumber) BUT NOT BOTH. // it can not get Integer value henceforth. //note, it would have already dealt with circumstance of processing all //digits in both uniform numbers above in first loop in addition method. try { //checks if the total is greater than 10. This is the same index on number1, number2 and also //carry forward 1 if applicable if ((Integer.valueOf(lastDigitFirstNumber) + Integer.valueOf(lastDigitSecondNumber) + firstDigitTotalDigits)>=10) { total = Integer.valueOf(lastDigitFirstNumber) + Integer.valueOf(lastDigitSecondNumber) + firstDigitTotalDigits; System.out.println("\nOVER 10"); System.out.println("addition of: " + Integer.valueOf(lastDigitFirstNumber) + " + " + Integer.valueOf(lastDigitSecondNumber)); System.out.println("this is current total: " + total); //first digit carried forward firstDigitTotalDigits = (int)total/10; System.out.println("This is first digit, it will be carried forward: " + firstDigitTotalDigits); remainder=true; //this is stored in the String lastDigitTotalDigits=total%10; //it has to be stored on the most left hand side like real life grandTotal = Integer.toString(lastDigitTotalDigits) + grandTotal; System.out.println("running grand total: " + grandTotal); System.out.println("length first: " + lengthFirstNumber); System.out.println("length second: " + lengthSecondNumber); //try //{ addition(number1.substring(0, (lengthFirstNumber-1)), number2.substring(0, (lengthSecondNumber-1))); //} //catch (ArrayIndexOutOfBoundsException e) //{ // System.out.println("one of the numbers is shorter than other"); // System.exit(0); //} //need to worry about carry forward in recursive call } else { //we expect total to be a single digit total = Integer.valueOf(lastDigitFirstNumber) + Integer.valueOf(lastDigitSecondNumber) + firstDigitTotalDigits; grandTotal = Integer.toString(total) + grandTotal; System.out.println("\nless than 10"); System.out.println("addition of: " + Integer.valueOf(lastDigitFirstNumber) + " + " + Integer.valueOf(lastDigitSecondNumber)); System.out.println("this is current total: " + total); System.out.println("running grandtotal: " + grandTotal); System.out.println("length first: " + lengthFirstNumber); System.out.println("length second: " + lengthSecondNumber); //it needs to set the carry forward value to 0 firstDigitTotalDigits=0; //try //{ //we now pass the exact starting number, but trim last number off addition(number1.substring(0, (lengthFirstNumber-1)), number2.substring(0, (lengthSecondNumber-1))); //} //catch (ArrayIndexOutOfBoundsException e) //{ //} } } catch (NumberFormatException e) { System.out.println("123This is last digit first number:" + lastDigitFirstNumber); System.out.println("123This is last digit second number:" + lastDigitSecondNumber); //now in here, it has to check if there is a carry over of 1 //if so it would add it to last digit of either number1 or number2 //for initial numbers such as, it is fortunately relatively more straight forward //4636 //+ 524 //but it was in a try and catch since if any of the values: //lastDigitFirstNumber="" or lastDigitSecondNumber="" //it would cause NumberFormatException since it can not format String => int //if the String is empty! We know however ONLY one of them is empty //if ((Integer.valueOf(lastDigitFirstNumber) + Integer.valueOf(lastDigitSecondNumber) //the alternate technique would be to add 0's on the front of the shortest number //to ensure both have uniform length, however this would affect readability when performing //System.out.println() // this is fine here System.out.println("****************"); System.out.println("number 1 at this point: " + number1); System.out.println("number 2 at this point: " + number2); //there are lots options in the if statement //can check if number2.length()>0 //can check if Integer.valueOf(number2)>0 //but best to examine lastDigitSecondNumber since more relevant to //leading most significant numbers... try { if (lastDigitSecondNumber!=null) { total = Integer.valueOf(lastDigitFirstNumber) + firstDigitTotalDigits; } //at this point we have dealt with scenarios such as these since it has logic //to carry forward 1 // 4636 // + 324 // 4636 // + 524 //but if the numbers were: 84636 // + 524 //it would fail since it has not had opportunity to process the 8 in number1 //so simply, once it has performed total with 4 and no value in number2 //it needs to take all the remaining numbers in the substring //and append it to the grand total //in effect it will shorten length number again by 1.. //note this time the shortening is normal. remainingPortionInteger = (int) (Integer.valueOf(number1) / 10); remainingPortionString = Integer.toString(remainingPortionInteger); System.out.println("1This is remaining portion after conversion: " + remainingPortionString); System.out.println("length of first number: " + lengthFirstNumber); remainingPortion = remainingPortionString.substring(0, (lengthFirstNumber-1)); System.out.println("This is remaining part: " + remainingPortion); } catch (NumberFormatException n) { total = Integer.valueOf(lastDigitSecondNumber) + firstDigitTotalDigits; //As above 524 //but if the numbers were: + 84636 // //it would fail since it has not had opportunity to process the 8 in number2 //but first it has to perform /10 to exclude the 4 remainingPortionInteger = (int) (Integer.valueOf(number2) / 10); remainingPortionString = Integer.toString(remainingPortionInteger); remainingPortion = remainingPortionString.substring(0, (lengthSecondNumber-1)); System.out.println("This is remaining portion after conversion: " + remainingPortionString); System.out.println("This is remaining part: " + remainingPortion); } //this part needs addressing. grandTotal = remainingPortion + Integer.toString(total) + grandTotal; System.out.println(grandTotal); return grandTotal; //System.exit(0); } return "test"; //} // end of if number%10!=0 } }