ProgrammingFundamentals

Add Two Numbers

Description

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 zeros, except the number 0 itself.

Example(s)

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

Analysis

Provided 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.

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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 LL1 and LL2. Since each digit is in the range of 0...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 the current node to a dummy head of the returning list.
• Initialize carry to 0.
• Initialize p and q to head of LL1 and LL2 respectively.
• Loop through lists LL1 and LL2 until you reach both ends.
• Set x to node p’s value. If p has reached the end of LL1, set to 0.
• Set y to node q’s value. If q has reached the end of LL2, set to 0.
• Set sum = x + y + carry.
• Update carry = sum / 10.
• Create a new node with the digit value of (sum mod 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 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:

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.

Java Code:

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

Complexity Analysis

Time complexity : O(max(m,n)). Assume that mm and nn represents the length of l1 and l2 respectively, the algorithm above iterates at most max(m,n) times.

Space complexity : O(max(m,n)). The length of the new list is at most max(m,n)+1.

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

Coded Solution

We can create a loop that goes through an element from both of the provided inputs, starting from the first/root element & end the loop when both the element’s “next” node points to a null. In each iteration of the loop, we can add the values & set that to the result node (if it’s the first iteration) or set it to the result’s next node in the list.

Complexity Analysis

Time: Since we go over both the lists once, the time complexity will be O(n) where n is the larger of the two lists (in terms of the number of elements). Space: We will require constant space for the temporary variables, but the space required for the result will increase with the inputs.

Statistics

On 04/03/2019:

Runtime: 19 ms, faster than 99.90% of Java online submissions for Add Two Numbers. Memory Usage: 47.8 MB, less than 41.96% of Java online submissions for Add Two Numbers.

Solution

• Java

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