Linked List

Linked List

1. Add Two Numbers: leetcode.cn/problems/add-two-numbers/

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.

Approach and Solution (Linked List + Carry)

1. Use a “dummy head node” head, and let l3 point to the tail of the current result list to facilitate inserting new nodes uniformly.
2. Use a variable s to act as the “carry” (in each round, first add the current digits to get the new s, then use s % 10 for the current digit and s // 10 to update the carry).
3. The loop condition is l1 or l2 or s: continue if either linked list has unprocessed digits or if there is still a carry.
4. Value retrieval in each round: treat empty lists as 0; after constructing the new node, move the pointers forward.

# 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: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        head = l3 = ListNode(0)
        s = 0
        while l1 or l2 or s:
            s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + s
            l3.next = ListNode(s % 10)
            s = s // 10
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
            l3 = l3.next 
        return head.next

Complexity Analysis

• Time Complexity: O(max(m, n)), iterating through both linked lists digit by digit once.
• Space Complexity: O(max(m, n)), the length of the result linked list is at most 1 digit longer than the longer input list (due to carry).