add 2 numbers

2. add 2 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

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解：

对我来说是学习Python里“链表”的一道题。
https://leetcode.com/articles/add-two-numbers/

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

首先要keep track of carry, 就是进位。
dummy是我们最后要返回的class。
v1 = v2 = 0 初始为零，确保了当一个list结束之后不会有问题,会直接表示value是0。
其中divmod 函数能够直接获得carry和val，于是新建一个ListNode，让“指针”指向下一个值。

class Solution(object):    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        carry = 0        dummy = n = ListNode(0)        while l1 or l2 or carry:            v1 = v2 = 0            if l1:                v1 = l1.val                l1 = l1.next            if l2:                v2 = l2.val                l2 = l2.next            carry, val = divmod(v1 + v2 + carry, 10)            n.next = ListNode(val)            n = n.next        return dummy.next