LeetCode（Add Two Numbers）

一、题目要求

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、解法

C语言  

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     struct ListNode *next;
  * };
  */
 struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
     struct ListNode* p1 = l1, *p2 = l2;
     struct ListNode* new_node = ;
     struct ListNode* ret_node = ;
     struct ListNode* temp_node = ;
     int sum = ;

     while(p1 != NULL || p2 != NULL || sum != ) {
         int temp = (p1 ? p1->val : ) + (p2 ? p2->val : ) + sum;
         int val = temp % ;
         sum = temp / ;

         new_node = (struct ListNode*)malloc(sizeof(struct ListNode));
         new_node->val = val;
         new_node->next = ;

         if ( == temp_node) {
             temp_node = new_node;
             ret_node = temp_node;
         } else {
             temp_node->next = new_node;
             temp_node = new_node;
         }

         p1 = p1 ? p1->next : p1;
         p2 = p2 ? p2->next : p2;
     }
     return ret_node;
 }

分析：根据题目的要求，将链表中对应元素相加，最后，倒序输出为一个链表结构。

需要注意：

• 链表长度不相同情况，比如一个[1,2,3]，例外一个是[1,2,3,4,5,6]情况时，实现方法：

  p1 ? p1->val : 

• 考虑进位情况，实现方法：

  int temp = (p1 ? p1->val : ) + (p2 ? p2->val : ) + sum;
  int val = temp % ;
  sum = temp / ;

• 考虑输出链表顺序问题，实现方法：

  if ( == temp_node) {
      temp_node = new_node;
      ret_node = temp_node;
  } else {
      temp_node->next = new_node;
      temp_node = new_node;
  }

运行结果：

Python  

 # Definition for singly-linked list.
 # class ListNode(object):
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None

 class Solution(object):
     def addTwoNumbers(self, l1, l2):
         """
         :type l1: ListNode
         :type l2: ListNode
         :rtype: ListNode
         """
         head = ListNode(0)
         temp = head
         carry = 0

         while l1 or l2 or carry:
             v1 = 0;
             v2 = 0;
             if l1:
                 v1 = l1.val
                 l1 = l1.next
             if l2:
                 v2 = l2.val
                 l2 = l2.next
             carry, v = divmod(v1 + v2 + carry, 10)
             temp.next = ListNode(v)
             temp = temp.next

         return head.next

运行结果：