Add Two Numbers

今天的题目是两数相加。

2.Add Two 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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

Explanation: 342 + 465 = 807.

2. 两数相加

给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)

输出：7 -> 0 -> 8

原因：342 + 465 = 807

My answer:

首先创建两个指针指向结果链表的头节点，一个指针dummy始终指在头节点，一个指针now用来指向尾结点(新值插入的位置)。然后设置一个进位标志carry初始化为0。x来代表l1的数值，y代表l2数值，任意一个链表的结束时其对应数值设为0，直到两个链表均结束循环停止。然后在循环内，获得当前位的值sum = x+y+carry和进位carry = sum//10，并将新值sum%10接在now指针后面。最后循环结束时，判断是否依然有进位，如果有进位则在结果链表后新增值为1的结点即可。最后返回dummy.next(注意返回时略过头节点)链表。

Runtime: 40 ms, faster than 99.89% of Python online submissions for Add Two Numbers.

     Memory Usage: 11.9 MB, less than 31.51% of Python online submissions for Add Two Numbers.

# 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
        """
        now = dummy = ListNode(0) #new node 
        carry = 0
        while(l1 or l2):
            x = l1.val if l1 is not None else 0
            y = l2.val if l2 is not None else 0
            sum = x+y+carry
            carry = sum//10
            now.next = ListNode(sum%10)
            now = now.next
            if(l1): l1 = l1.next
            if(l2): l2 = l2.next
        if(carry):
            now.next = ListNode(1)
        return dummy.next