Leetcode and Lintcode

Add Two Numbers II

Problem:

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in forward order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

Lintcode link

Example
Given 6->1->7 + 2->9->5. That is, 617 + 295.

Return 9->1->2. That is, 912.

Solution:

Revese the given two linked list.
Then use the solution presented in the solution of “Add Two Numbers” to add then together.
Finally, reverse the result linked list back.

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/*************************************************************************
	> File Name: AddTwoNumbersII.java
	> Author: Yao Zhang 
	> Mail: psyyz10@163.com 
	> Created Time: Mon 16 Nov 14:27:38 2015
 ************************************************************************/

public class AddTwoNumbersII{
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    public ListNode addLists2(ListNode l1, ListNode l2) {
        l1 = reverse(l1);
        l2 = reverse(l2);

        return reverse(addLists1(l1,l2));
    }

    public ListNode reverse(ListNode l){
        ListNode head = null;

        while (l != null){
            ListNode temp = l.next;
            l.next = head;
            head = l;
            l = temp;
        }

        return head;
    }

    public ListNode addLists1(ListNode l1, ListNode l2){
        if (l1 == null && l2 == null)
            return null;

        ListNode head = new ListNode(0);
        ListNode current = head;
        int carry = 0;

        while (l1 != null || l2 != null){
            int sum = carry;

            if (l1 != null){
                sum += l1.val;
                l1 = l1.next;
            }

            if (l2 != null){
                sum += l2.val;
                l2 = l2.next;
            }

            carry = sum / 10;
            current.next = new ListNode(sum % 10);
            current = current.next;
        }

        if (carry != 0){
            current.next = new ListNode(carry);
        }

        return head.next;
    }
}