2. Add Two Numbers
1. Description
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.
2. Example
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]
3. 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.
4. Solutions
My Accepted Solution
n = max(length of l1, length of l2)
Time complexity: O(n)
Space complexity: O(n)
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode head;
ListNode *iter = &head;
int last_carry = 0, carry = 0;
while (l1 != nullptr && l2 != nullptr) {
carry = (l1->val + l2->val + last_carry) / 10;
int digit = (l1->val + l2->val + last_carry) - carry * 10;
last_carry = carry;
ListNode* node = new ListNode(digit);
iter->next = node;
iter = iter->next;
l1 = l1->next;
l2 = l2->next;
}
ListNode *left_iter = (l1 == nullptr ? l2 : l1);
while (left_iter != nullptr) {
carry = (left_iter->val + last_carry) / 10;
int digit = (left_iter->val + last_carry) - carry * 10;
last_carry = carry;
ListNode* node = new ListNode(digit);
iter->next = node;
iter = iter->next;
left_iter = left_iter->next;
}
if (carry == 1) {
ListNode* node = new ListNode(1);
iter->next = node;
}
return head.next;
}
};