一、題目要求

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語言

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     struct ListNode *next;
 6  * };
 7  */
 8 struct ListNode* addTwoNumbers(struct
 
 ListNode* l1, struct ListNode* l2) {
 9     struct ListNode* p1 = l1, *p2 = l2;
10     struct ListNode* new_node = 0;
11     struct ListNode* ret_node = 0;
12     struct ListNode* temp_node = 0;
13     int sum = 0;
14 
15     while(p1 != NULL || p2 != NULL || sum != 0) {
16         int temp = (p1 ? p1->val : 0
 
) + (p2 ? p2->val : 0) + sum;
17         int val = temp % 10;
18         sum = temp / 10;
19 
20         new_node = (struct ListNode*)malloc(sizeof(struct ListNode));
21         new_node->val = val;
22         new_node->next = 0;
23 
24         if (0 == temp_node) {
25             temp_node = new_node;
26             ret_node = temp_node;
27         } else {
28             temp_node->next = new_node;
29             temp_node = new_node;
30         }
31 
32         p1 = p1 ? p1->next : p1;
33         p2 = p2 ? p2->next : p2;
34     }
35     return ret_node;
36 }

分析：根據題目的要求，將連結串列中對應元素相加，最後，倒序輸出為一個連結串列結構。

需要注意：

• 連結串列長度不相同情況，比如一個[1,2,3]，例外一個是[1,2,3,4,5,6]情況時，實現方法：

  p1 ? p1->val : 0

• 考慮進位情況，實現方法：

  int temp = (p1 ? p1->val : 0) + (p2 ? p2->val : 0) + sum;
  int val = temp % 10;
  sum = temp / 10;

• 考慮輸出連結串列順序問題，實現方法：

  if (0 == temp_node) {
      temp_node = new_node;
      ret_node = temp_node;
  } else {
      temp_node->next = new_node;
      temp_node = new_node;
  }

   

執行結果：