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Leetcode - 2. Add Two Numbers

Understanding Carry in Linked List Addition Key Concepts Numbers are stored in reverse order as linked lists. Each node contains a single digit. Carry is needed when sum > 9. Result is returned as a new linked list. Approach Initialize a dummy node to simplify list construction. Use a carry variable to handle sums exceeding 9. Traverse l1 and l2, summing corresponding nodes. If one list is shorter, treat missing nodes as 0. Continue processing even if one list is exhausted. Add a new node if there’s a leftover carry. Complexity Analysis Time Complexity: O(max(N, M)) (where N, M are the lengths of l1 and l2). Space Complexity: O(max(N, M)) (for the new linked list). Code Implementation var addTwoNumbers = function (l1, l2) { let dummy = new ListNode(0); let current = dummy; let carry = 0; while (l1 !== null || l2 !== null || carry > 0) { let sum = (l1 ? l1.val : 0) + (l2 ? l2.val : 0) + carry; carry = Math.floor(sum / 10); current.next = new ListNode(sum % 10); current = current.next; if (l1) l1 = l1.next; if (l2) l2 = l2.next; } return dummy.next; };

Mar 21, 2025 - 19:10

Understanding Carry in Linked List Addition

Key Concepts

Numbers are stored in reverse order as linked lists.
Each node contains a single digit.
Carry is needed when sum > 9.
Result is returned as a new linked list.

Approach

Initialize a dummy node to simplify list construction.
Use a carry variable to handle sums exceeding 9.
Traverse l1 and l2, summing corresponding nodes.
If one list is shorter, treat missing nodes as 0.
Continue processing even if one list is exhausted.
Add a new node if there’s a leftover carry.

Complexity Analysis

Time Complexity: O(max(N, M)) (where N, M are the lengths of l1 and l2).
Space Complexity: O(max(N, M)) (for the new linked list).

Code Implementation

var addTwoNumbers = function (l1, l2) {
    let dummy = new ListNode(0);
    let current = dummy;
    let carry = 0;

    while (l1 !== null || l2 !== null || carry > 0) {
        let sum = (l1 ? l1.val : 0) + (l2 ? l2.val : 0) + carry;
        carry = Math.floor(sum / 10);
        current.next = new ListNode(sum % 10);
        current = current.next;

        if (l1) l1 = l1.next;
        if (l2) l2 = l2.next;
    }
    return dummy.next;
};