algorithm to do arbitrary precision addition using linked lists:


L1: list addend
L2: list addend
L3: list sum


finds out which list has more digits (list.size())

L1_index= L1.size()-1
L2_index= L2.size()-1

make list L2 the longer of the two lists - swap if necessary

carry=0;
while L1_index>=0 { //L1 is shorter

   new_digit= L1[L1_index] + L2[L2_index] + carry
   if new_digit<10 insert new_digit in sum, set carry =0
      else insert (new_digit%10) in sum, set carry=1
   
   Decrement L1_index
   Decrement L2_index
}

4 cases now:

L2_index<0, carry=0: do nothing
L2_index<0, carry=1: insert 1 in sum list.
L2_index>=0, carry=0: insert remaining digits in L2 into sum
L2_index>=0, carry=1: let x=next digit in L2 + carry.  If x is <10
            insert in sum, set carry=0, and insert remaining digits of
	    L2 into sum.

            if x>=10, insert x%10 into sum, and repeat for next digit
	    with carry =1.

            When L2_index<0,if carry still = 1, insert 1 into result.