# Problem

problem at: https://leetcode.com/problems/add-two-numbers-ii

See the easier related problem Add Two Numbers.

# Solution

Among the following solutions, 1st solution is easiest to think of but need to know how to reverse a singly linked list by heart. 2nd solution could be thought of when LIFO rings a bell. last solution could be thought of but takes time to implement.

# Reverse Linked List & DIY ListNode Adder

Create a nested function reverseList to reverse the two input linked list and then construct a ListNode adder, similar as before.


  • Time:
  • Space: due to no extra space

where m is the number of nodes in l1 and n is the number of nodes in l2.

def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    def reverseList(l: ListNode) -> ListNode:
        prev = None
        curr = l

        while curr:
            # IMPT: remember the steps to reverse a singly linked list!!!
            nxt = curr.next
            curr.next = prev
            prev = curr
            curr = nxt
        return prev

    new_l1 = reverseList(l1)
    new_l2 = reverseList(l2)

    carry = 0
    end = dummy = ListNode(0)
    while new_l1 or new_l2 or carry:
        if new_l1:
            carry += new_l1.val
            new_l1 = new_l1.next
        if new_l2:
            carry += new_l2.val
            new_l2 = new_l2.next
        end.next = ListNode(carry%10)
        end = end.next
        carry = carry//10

    return reverseList(dummy.next)

# Follow Up

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

There are two ways: stack and recursion.

# Stack

Initialize two stacks for l1 and l2 respectively: l1_stack and l2_stack. After appending, I then pop each stack and calculate the sum. Note that now sum is also reversed, so the way to construct is different from previous question.


  • Time:
  • Space: due to extra space of 2 stacks
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    l1_stack = []
    l2_stack = []

    while l1:
        l1 = l1.next
    while l2:
        l2 = l2.next

    l1_len = len(l1_stack)
    l2_len = len(l2_stack)

    head = temp = None
    i = j = 0
    carry = 0

    while i<l1_len or j<l2_len or carry:
        if i<l1_len:
            carry += l1_stack.pop()
            i += 1
        if j<l2_len:
            carry += l2_stack.pop()
            j += 1
        temp = ListNode(carry%10)
        temp.next = head
        head = temp
        carry //= 10

    return head

# Recursion (REDO!)

Frankly, I copied most of the following code, as the recursive function add_list is not easy to write(especially what vars to return). The idea is:

  1. get the difference in length of l1 and l2
  2. IMPT: recursively add two lists and return head and new_carry. Note that diff decreases only when ln1 is longer than ln2
  3. if there is a leftmost carry, make it head and return


  • Time:
  • Space: due to implicit stack space
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    def get_len(l: ListNode) -> int:
        length = 0
        while l:
            l = l.next
            length += 1
        return length

    def add_lists(ln1: ListNode, ln2: ListNode, diff: int) -> (ListNode, int):
        if ln1 is None and ln2 is None:
            return None, 0
        if diff>0:
            # currently ln1 is longer than ln2
            # move the pointer at list1 to n1.next, don't move the pointer at list2
            next_node, carry = add_lists(ln1.next, ln2, diff-1)
            carry += ln1.val
            next_node, carry = add_lists(ln1.next, ln2.next, diff)
            carry += ln1.val + ln2.val
        new_val, new_carry = carry%10, carry//10
        head = ListNode(new_val)
        head.next = next_node
        return head, new_carry

    l1_len, l2_len = get_len(l1), get_len(l2)
    # always keep len(ln1) > len(ln2)
    if l1_len < l2_len:
        l1_len, l2_len = l2_len, l1_len
        l1, l2 = l2, l1
    diff = l1_len - l2_len
    head, carry = add_lists(l1, l2, diff)
    # handle the leftmost carry
    if carry:
        c = ListNode(carry)
        c.next = head
        head = c
    return head