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 contains a single digit. Add the two numbers and return the sum as a linked list.

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

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Constraints:

  • The number of nodes in each linked list is in the range [1, 100].
  • 0 <= Node.val <= 9
  • It is guaranteed that the list represents a number that does not have leading zeros.

Approach 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.

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 l1l1 and l2l2. Since each digit is in the range of 0 \ldots 90…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 = 1carry=1 to the next iteration. carrycarry must be either 00 or 11 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 00.
  • Initialize pp and qq to head of l1l1 and l2l2 respectively.
  • Loop through lists l1l1 and l2l2 until you reach both ends.
  • Set xx to node pp’s value. If pp has reached the end of l1l1, set to 00.
  • Set yy to node qq’s value. If qq has reached the end of l2l2, set to 00.
  • Set sum = x + y + carrysum=x+y+carry.
  • Update carry = sum / 10carry=sum/10.
  • Create a new node with the digit value of (sum \bmod 10)(summod10) and set it to current node’s next, then advance current node to next.
  • Advance both pp and qq.
  • Check if carry = 1carry=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:

Complexity Analysis

  • Time complexity : O(max(m,n)). Assume that mm and n 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

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