2. Add Two Numbers

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2. Add Two Numbers

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 sumas a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

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Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

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Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

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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.

Hints/Notes

Solution

Language: C++

Recursive:

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2, int prev = 0) {
        if (!l1 && !l2) {
            return prev ? new ListNode(prev) : nullptr;
        }
        if (!l1) {
            swap(l1, l2);
        }
        int sum = l1->val + (l2 ? l2->val : 0) + prev;
        ListNode* node = new ListNode(sum % 10);
        node->next = addTwoNumbers(l1->next, l2 ? l2->next : l2, sum / 10);
        return node;
    }
};

Iterative:

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode dummy(0);
        ListNode* head = &dummy;
        int prev = 0;
        while (l1 || l2 || prev) {
            if (l1) {
                prev += l1->val;
                l1 = l1->next;
            }
            if (l2) {
                prev += l2->val;
                l2 = l2->next;
            }
            head->next = new ListNode(prev % 10);
            prev /= 10;
            head = head->next;
        }
        return dummy.next;
    }
};