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Question
Solution

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

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Solution

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: $342 + 465 = 807$ .
\nEach 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 $0 \\ldots 9$ , summing two digits may "overflow". For example $5 + 7 = 12$ . In this case, we set the current digit to $2$ 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 $9 + 9 + 1 = 19$ .

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 and 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 $(sum \\bmod 10)$ and set it to current node\'s next, then advance current node to next.
- Advance both $p$ and $q$ .
\n
Check if $carry = 1$ , if so append a new node with digit $1$ 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:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Test case	Explanation
\n $l1=[0,1]$ $l2=[0,1,2]$ \n	When one list is longer than the other.
\n $l1=[]$ $l2=[0,1]$ \n	When one list is null, which means an empty list.
\n $l1=[9,9]$ $l2=[1]$ \n	The sum could have an extra carry of one at the end, which is easy to forget.

\n\n

Complexity Analysis

\n
Time complexity : $O(\\max(m, n))$ . Assume that $m$ and $n$ represents the length of $l1$ and $l2$ respectively, the algorithm above iterates at most $\\max(m, n)$ times.
\n
\n
Space complexity : $O(\\max(m, n))$ . The length of the new list is at most $\\max(m,n) + 1$ .
\n

Follow up

What if the the digits in the linked list are stored in non-reversed order? For example:

\n $\n(3 \\to 4 \\to 2) + (4 \\to 6 \\to 5) = 8 \\to 0 \\to 7\n$ \n

2. Add Two Numbers

Solution