[LeetCode]2. Add Two Numbers_Medium tag: Linked List
2021-07-27 22:00 Johnson_强生仔仔 阅读(10) 评论(0) 编辑 收藏 举报
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.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0] Output: [0]
Example 3:
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.
Ideas: use a pre (init: 0) to add into the curSum. Rember to check the pre after adding everything, because assume 9 + 1 -> 10. the 1 is saved in pre.
T: O(m + n), S: O(1) m: len(l1), n: len(l2)
Code:
dummy = ListNode() head = dummy pre = 0 while l1 or l2: curSum = 0 if l1: curSum += l1.val l1 = l1.next if l2: curSum += l2.val l2 = l2.next curSum += pre pre, rem = divmod(curSum, 10) newNode = ListNode(rem) head.next = newNode head = head.next if pre: head.next = ListNode(pre) return dummy.next