Algorithm——Add Two Numbers（补上周）

一、question

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

二、solution

（cpp）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode preHead(0), *p = &preHead;
    int carry=0;
        while(l1||l2||carry){
            int sum = (l1?l1->val:0)+(l2?l2->val:0)+carry;
            carry = sum/10;
            p->next = new ListNode(sum % 10);
            p = p->next;
            l1 = l1?l1->next:0;
            l2 = l2?l2->next:0;
        }
        return preHead.next;    
    }
};

（js）

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */

var addTwoNumbers = function(l1, l2) {
  var add = 0
    , ans
    , head;

  while(l1 || l2) {
    var a = l1 ? l1.val : 0
      , b = l2 ? l2.val : 0;

    var sum = a + b + add;
    add = ~~(sum / 10);

    var node = new ListNode(sum % 10);

    if (!ans)
      ans = head = node;
    else {
      head.next = node;
      head = node; 
    }
    
    if (l1)
      l1 = l1.next;
    if (l2)
      l2 = l2.next;
  }

  if (add) {
    var node = new ListNode(add);
    head.next = node;
    head = node;
  }

  return ans;
};