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: ( ->  -> ) + ( ->  -> )

Output:  ->  ->

Explanation:  +  = .

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(), *p = &preHead;

    int carry=;

        while(l1||l2||carry){

            int sum = (l1?l1->val:)+(l2?l2->val:)+carry;

            carry = sum/;

            p->next = new ListNode(sum % );

            p = p->next;

            l1 = l1?l1->next:;

            l2 = l2?l2->next:;

        }

        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;

};

巴特西

Algorithm——Add Two Numbers（补上周）

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