Add two numbers

Time: O(N); Space: O(1); medium

You are given two non-empty linked lists representing two non-negative integers.

The digits are stored in reverse order and each of their nodes contain a single digit.

Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = {ListNode} 2->4->3->None, l2 = {ListNode} 5->6->4->None

Output: {ListNode} 7->0->8->None

Explanation:

  • 342 + 465 = 807

Follow up:

  • What if the the digits in the linked list are stored in non-reversed order?

  • For example: 3->4->2 + 4->6->5 = 8->0->7

[8]:

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

1. Elementary Math

Intuition

Keep track of the carry using a variable and simulate digits-by-digits sum starting from the head of list, which contains the least-significant digit.

Illustration of Adding two numbers

Figure 1. Visualization of the addition of two numbers: 342 + 465 = 807. Each node contains a single digit and the digits are stored in reverse order.

Algorithm

Just like how you would sum two numbers on a piece of paper, we begin by summing the least-significant digits, which is the head of l1 and l2. Since each digit is in the range of 0…9, summing two digits may “overflow”.

For example 5 + 7 = 125 + 7 = 12. In this case, we set the current digit to 22 and bring over the carry = 1 to the next iteration. carry must be either 0 or 1 because the largest possible sum of two digits (including the carry) is 9 + 9 + 1 = 199 + 9 + 1 = 19.

The pseudocode is as following:

  • Initialize current node to dummy head of the returning list.

  • Initialize carry to 0.

  • Initialize p and q to head of l1 and l2 respectively.

  • Loop through lists l1 and l2 until you reach both ends.

    • Set x to node p’s value. If p has reached the end of l1, set to 0.

    • Set y to node q’s value. If q has reached the end of l2, set to 0.

    • Set sum = x + y + carry.

    • Update carry = sum // 10.

    • Create a new node with the digit value of sum % 10 and set it to current node’s next, then advance current node to next.

    • Advance both p and q.

  • Check if carry = 1, if so append a new node with digit 11 to the returning list.

  • Return dummy head’s next node.

Note that we use a dummy head to simplify the code. Without a dummy head, you would have to write extra conditional statements to initialize the head’s value.

Take extra caution of the following cases:

Test case

Explanation

l1 = [0,1], l2=[0,1,2]

When one list is longer than the other

l1=[], l2=[0,1]

When one list is null, which means an empty list

l1=[9,9], l2=[1]

The sum could have an extra carry of one at the end, which is easy to forget

 
[9]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(1)
    """
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(0)
        current, carry = dummy, 0

        while l1 or l2:
            val = carry
            if l1:
                val += l1.val
                l1 = l1.next
            if l2:
                val += l2.val
                l2 = l2.next

            carry, val = divmod(val, 10)
            current.next = ListNode(val)
            current = current.next

        if carry == 1:
            current.next = ListNode(1)

        return dummy.next

[10]:

s = Solution1()

l1 = ListNode(2)
l1.next = ListNode(4)
l1.next.next = ListNode(3)
l2 = ListNode(5)
l2.next = ListNode(6)
l2.next.next = ListNode(4)
res = s.addTwoNumbers(l1, l2)
exp = [7,0,8]
for val in exp:
    assert res.val == val
    res = res.next

l1 = ListNode(3)
l1.next = ListNode(4)
l1.next.next = ListNode(2)
l2 = ListNode(4)
l2.next = ListNode(6)
l2.next.next = ListNode(5)
res = s.addTwoNumbers(l1, l2)
exp = [7,0,8]
for val in exp:
    assert res.val == val
    res = res.next