[LC] 2. Add Two NumbersTech
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 = , l2 =  Output: 
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] Output: [8,9,9,9,0,0,0,1]
- The number of nodes in each linked list is in the range
0 <= Node.val <= 9
- It is guaranteed that the list represents a number that does not have leading zeros.
Approach is simple. Just looking all of the list pair of list1 and list2, and calculate sum = (list1.val + list2.val + prev_carry)%10 and carry = (list1.val + list2.val + prev_carry) // 10. Loop both list first, and list1, list2 consequentially. If carry is 1 at the last, we should add it to the last element.
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: root = ListNode(0) head = root s,c = 0,0 while l1 and l2: s,c = (l1.val+l2.val+c)%10, (l1.val+l2.val+c)//10 root.next = ListNode(s) l1,l2,root = l1.next, l2.next, root.next while l1: s,c = (l1.val+c)%10, (l1.val+c)//10 root.next = ListNode(s) l1, root = l1.next, root.next while l2: s,c = (l2.val+c)%10, (l2.val+c)//10 root.next = ListNode(s) l2, root = l2.next, root.next if c != 0: root.next = ListNode(c) return head.next
Time/Space Complexity: O(N)
Runtime: 88 ms, faster than 10.65% of Python3 online submissions for Add Two Numbers.
Memory Usage: 14.4 MB, less than 12.54% of Python3 online submissions for Add Two Numbers.