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 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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

Examples

Example 1

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

Explanation: 342 + 465 = 807.

Example 2

Input: (9 -> 9 -> 9 -> 9 -> 9 -> 9 -> 9) + (9 -> 9 -> 9 -> 9)

Output: 8 -> 9 -> 9 -> 9 -> 0 -> 0 -> 0 -> 1

Explanation: (9999999) + (9999) = 89990001

Solutions

Solution 1

/**
 * 
 * 
 */

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode rootNode = null;
        int carry = 0;
        while(l1 != null || l2 != null) {
            int sum = getValue(l1) + getValue(l2) + carry;
            if(sum / 10 > 0) {
                carry = sum / 10;
                sum = sum % 10;
            } else {
                carry = 0;
            }
            rootNode = insertNode(rootNode, sum);
            if(l1 != null) {
                l1 = l1.next;
            }
            if(l2 != null) {
                l2 = l2.next;
            }
        }
        if(carry > 0) {
            rootNode = insertNode(rootNode, carry);
        }
        return rootNode;
    }
    
    public int getValue(ListNode node) {
        if(node == null) {
            return 0;
        } else {
            return node.val;
        }
    }
    
    public ListNode insertNode(ListNode rootNode, int val) {
        ListNode newNode = new ListNode(val);
        if(rootNode == null) {
            rootNode = newNode;
        } else {
            ListNode tempNode = rootNode;
            while(tempNode.next != null) {
                tempNode = tempNode.next;
            }
            tempNode.next = newNode;
        }
        return rootNode;
    }
}

Solution 2

/**
 * Time complexity : O(max(m,n)). Assume that m and n represents the length of 
 *     l1 and l2 respectively, the algorithm above iterates at most max(m,n) times.
 * Space complexity : O(max(m,n)). The length of the new list is at most max(m,n)+1.
 */

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode root = new ListNode(0);
        ListNode curr = root;
        int carry = 0, s = 0;
        while(!(l1 == null && l2 == null)) {
            s = (l1 == null)? 0: l1.val;
            s += (l2 == null)? 0: l2.val;
            s += carry;
            carry = s/10;
            curr.next = new ListNode(s%10);
            curr = curr.next;
            if(l1 != null) l1 = l1.next;
            if(l2 != null) l2 = l2.next;
        }
        if(carry > 0) curr.next = new ListNode(carry);
        return root.next;
    }
}

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