Add Two Numbers II (Ladder)

Description

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.

Example
Given 6->1->7 + 2->9->5. That is, 617 + 295.
Return 9->1->2. That is, 912.

>

Link

Lintcode_ladder

Method

1. x
2. x

Example

1. 1

   /**
   * Definition for singly-linked list.
   * struct ListNode {
   *     int val;
   *     ListNode *next;
   *     ListNode(int x) : val(x), next(NULL) {}
   * };
   */
   class Solution {
   public:
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    ListNode *addLists2(ListNode *l1, ListNode *l2) {
        // write your code here
        if (!l1) {
            return l2;
        } else if (!l2) {
            return l1;
        }
        ListNode* rev1 = reverseList(l1);
        ListNode* rev2 = reverseList(l2);
        ListNode* sum = addtwo(rev1, rev2);
        return reverseList(sum);
    }
    // reverse list
    ListNode* reverseList(ListNode* head) {
        ListNode* tail = nullptr;
        while (head) {
            ListNode* next = head->next;
            head->next = tail;
            tail = head;
            head = next;
        }
        return tail;
    }
    // add two list with reversed list
    ListNode* addtwo(ListNode* left, ListNode* right) {
        ListNode dummy(0);
        ListNode* prev = &dummy;
        int sum = 0;
        int carry = 0;
        while (left || right) {
            // left && right are not empty
            if (left && right) {
                sum = left->val + right->val + carry;
                left = left->next;
                right = right->next;
            } else if (left) {
            // only left is not empty
                sum = left->val + carry;
                left = left->next;
            } else {
            // only right is not empty
                sum = right->val + carry;
                right = right->next;
            }
            // update add result
            carry = sum / 10;
            sum %= 10;
            prev->next = new ListNode(sum);
            prev = prev->next;
        }
        if (carry > 0) {
            prev->next = new ListNode(carry);
        }
        return dummy.next;
    }
   };
   

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