leetcode:2-Add Two Numbers-优快云博客

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.

题意：给定两个非空链表来表示两个非负整数。数字以相反的顺序存储，每个节点包含一个数字。对这两个数字加和并将其以链表形式返回。假定这两个数字不包含任何前导零，除了数字0本身。

思路：若存在空链表，直接返回另一个即可。若均不为空，就对应位置逐个相加。注意，提前实例化一个存放结果链表的头结点。

 1 class ListNode:
 2     def __init__(self,x):
 3         self.val=x
 4         self.next=None
 5 
 6 class Solution:
 7     def addTwoNumbers(self,l1,l2):
 8         if l1 is None:  #若存在空链表，直接返回另一个即可
 9             return l2
10         elif l2 is None:
11             return l1
12         else:
13             carry = 0   #进位标志
14             pNode = ListNode(0)  #存放结果的结点数值先放0
15             pHead = pNode        #指向第一个元素的指针
16             while(l1 or l2):  #直到某个表为空
17                 sum = 0  #移动指针后先清空
18                 if(l1):
19                     sum = l1.val  #sum存下l1当下指向的数
20                     l1 = l1.next  #存数后移动指针
21                 if(l2):
22                     sum += l2.val  #sum进行对应位置加法运算
23                     l2 = l2.next   #存数后移动指针
24                 sum += carry   #加上进位
25                 pHead.next = ListNode(sum%10)  #存下当前位的数字
26                 pHead = pHead.next  #指向该数字
27                 carry = (sum>=10)   #表示进位
28             if(carry):   #若结束l1和l2的遍历仍有进位
29                 pHead.next = ListNode(1)   #记下进位
30             pHead = pNode.next  #指向第一个位置
31             del pNode    #删除头结点
32             return pHead
33 
34 
35 if __name__=='__main__':
36     l1_1 = ListNode(2)
37     l1_2 = ListNode(4)
38     l1_3 = ListNode(3)
39     l1_1.next = l1_2
40     l1_2.next = l1_3
41     l2_1 = ListNode(5)
42     l2_2 = ListNode(6)
43     l2_3 = ListNode(4)
44     l2_1.next = l2_2
45     l2_2.next = l2_3
46     l3_1=Solution().addTwoNumbers(l1_1,l2_1)
47     while l3_1 != None:
48         print(l3_1.val)
49         l3_1 = l3_1.next