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.Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
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
