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 contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
Solution: Simulation
Time complexity: O(max(n,m))
Space complexity: O(max(n,m))
C++
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// Author: Huahua class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode dummy(0); ListNode* tail = &dummy; int sum = 0; while (l1 || l2 || sum) { sum += (l1 ? l1->val : 0) + (l2 ? l2->val : 0); l1 = l1 ? l1->next : nullptr; l2 = l2 ? l2->next : nullptr; tail->next = new ListNode(sum % 10); sum /= 10; tail = tail->next; } return dummy.next; } }; |
Java
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// Author: Huahua class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode tail = new ListNode(0); ListNode dummy = tail; int sum = 0; while (l1 != null || l2 != null || sum > 0) { sum += (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val); tail.next = new ListNode(sum % 10); tail = tail.next; if (l1 != null) l1 = l1.next; if (l2 != null) l2 = l2.next; sum /= 10; } return dummy.next; } } |
Python3
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# Author: Huahua class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: dummy = tail = ListNode(0) s = 0 while l1 or l2 or s: s += (l1.val if l1 else 0) + (l2.val if l2 else 0) tail.next = ListNode(s % 10) tail = tail.next s //= 10 l1 = l1.next if l1 else None l2 = l2.next if l2 else None return dummy.next |