LeetCode 445: Add Two Numbers II Solution

Master LeetCode problem 445 (Add Two Numbers II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

445. Add Two Numbers II

Medium

Linked List Math Stack

Problem Explanation

Explanation

To solve this problem without reversing the input lists, we can use a stack to store the values of the nodes in each linked list. By doing this, we can add the numbers starting from the least significant digit. We iterate through the lists while pushing their values onto the stacks. Then, we pop the values from the stacks and add them while keeping track of the carry. We create a new linked list to store the sum in reverse order. At the end, we reverse the result linked list to get the final sum in the correct order.

Algorithm:

Push the values of nodes in both linked lists onto two separate stacks.
Pop the values from the stacks while adding them and keeping track of the carry.
Create a new linked list to store the sum in reverse order.
Reverse the result linked list to obtain the final sum.

Time Complexity: O(max(m, n)) where m and n are the lengths of the two input linked lists.

Space Complexity: O(m + n) to store the values in the stacks and the result linked list.

Solution Code

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    Stack<Integer> stack1 = new Stack<>();
    Stack<Integer> stack2 = new Stack<>();
    
    while (l1 != null) {
        stack1.push(l1.val);
        l1 = l1.next;
    }
    while (l2 != null) {
        stack2.push(l2.val);
        l2 = l2.next;
    }
    
    ListNode dummy = new ListNode(0);
    int carry = 0;
    
    while (!stack1.isEmpty() || !stack2.isEmpty() || carry != 0) {
        int sum = carry;
        if (!stack1.isEmpty()) sum += stack1.pop();
        if (!stack2.isEmpty()) sum += stack2.pop();
        
        carry = sum / 10;
        sum %= 10;
        
        ListNode newNode = new ListNode(sum);
        newNode.next = dummy.next;
        dummy.next = newNode;
    }
    
    return dummy.next;
}

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