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
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 : 0 - 考虑进位情况，实现方法：
int temp = (p1 ? p1->val : 0) + (p2 ? p2->val : 0) + sum; int val = temp % 10; sum = temp / 10;
- 考虑输出链表顺序问题，实现方法：
if (0 == 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
1 # Definition for singly-linked list. 2 # class ListNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution(object): 8 def addTwoNumbers(self, l1, l2): 9 """ 10 :type l1: ListNode 11 :type l2: ListNode 12 :rtype: ListNode 13 """ 14 head = ListNode(0) 15 temp = head 16 carry = 0 17 18 while l1 or l2 or carry: 19 v1 = 0; 20 v2 = 0; 21 if l1: 22 v1 = l1.val 23 l1 = l1.next 24 if l2: 25 v2 = l2.val 26 l2 = l2.next 27 carry, v = divmod(v1 + v2 + carry, 10) 28 temp.next = ListNode(v) 29 temp = temp.next 30 31 return head.next
运行结果：
