LeetCode-2： Add Two Numbers

【Problem：2-Add Two Numbers】

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

【Solution】

1）From 九章算法：（can run, but "Submission Result： Wrong Answer！"）

 class Solution:
     def addTwoNumbers(self, l1, l2):
         head = ListNode(0)
         ptr = head
         carry  = 0
         while True:
             if l1 != None:
                 carry += l1.val
                 l1 = l1.next
             if l2 != None:
                 carry += l2.val
                 l2 = l2.next
             ptr.val = carry % 10
             carry /= 10
             # 运算未结束新建一个节点用于储存答案，否则退出循环
             if l1 != None or l2 != None or carry != 0:
                 ptr.next = ListNode(0)
                 ptr = ptr.next
             else:
                 break
         return head

2）Java ：

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         if(l1 == null && l2 == null) {
             return null;
         }

         ListNode head = new ListNode(0);
         ListNode point = head;
         int carry = 0;
         while(l1 != null && l2!=null){
             int sum = carry + l1.val + l2.val;
             point.next = new ListNode(sum % 10);
             carry = sum / 10;
             l1 = l1.next;
             l2 = l2.next;
             point = point.next;
         }

         while(l1 != null) {
             int sum =  carry + l1.val;
             point.next = new ListNode(sum % 10);
             carry = sum /10;
             l1 = l1.next;
             point = point.next;
         }

         while(l2 != null) {
             int sum =  carry + l2.val;
             point.next = new ListNode(sum % 10);
             carry = sum /10;
             l2 = l2.next;
             point = point.next;
         }

         if (carry != 0) {
             point.next = new ListNode(carry);
         }
         return head.next;
     }
 }

 // version: 高频题班
 public class Solution {
     /**
      * @param l1: the first list
      * @param l2: the second list
      * @return: the sum list of l1 and l2
      */
     public ListNode addLists(ListNode l1, ListNode l2) {
         // write your code here
         ListNode dummy = new ListNode(0);
         ListNode tail = dummy;

         int carry = 0;
         for (ListNode i = l1, j = l2; i != null || j != null; ) {
             int sum = carry;
             sum += (i != null) ? i.val : 0;
             sum += (j != null) ? j.val : 0;

             tail.next = new ListNode(sum % 10);
             tail = tail.next;

             carry = sum / 10;
             i = (i == null) ? i : i.next;
             j = (j == null) ? j : j.next;
         }

         if (carry != 0) {
             tail.next = new ListNode(carry);
         }
         return dummy.next;
     }
 }

3）C++：yes, accepted

/**
 * 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) {
        // 题意可以认为是实现高精度加法
        ListNode *head = new ListNode();
        ListNode *ptr = head;
        int carry = ;
        while (true) {
            if (l1 != NULL) {
                carry += l1->val;
                l1 = l1->next;
            }
            if (l2 != NULL) {
                carry += l2->val;
                l2 = l2->next;
            }
            ptr->val = carry % ;
            carry /= ;
            // 当两个表非空或者仍有进位时需要继续运算，否则退出循环
            if (l1 != NULL || l2 != NULL || carry != ) {
                ptr = (ptr->next = new ListNode());
            } else break;
        }
        return head;
    }
};

【References】

1、http://www.jiuzhang.com/solution/add-two-numbers/

Time complexity：O(n^2)2