原題：

https://leetcode.com/problems/add-two-numbers/description/

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.

分析：

明顯的鏈表問題。非負(fù)整數(shù)說明不用考慮符號(hào)和小數(shù)點(diǎn)，每個(gè)節(jié)點(diǎn)代表一位0-9的數(shù)字。并且輸入不存在0開頭的情況。
唯一需要注意的是輸入的兩個(gè)鏈表長(zhǎng)度可能不同，迭代循環(huán)需要小心。
再有進(jìn)位問題：即便兩個(gè)鏈表已經(jīng)算完，進(jìn)位上有值，仍然需要在和鏈表中生成新節(jié)點(diǎn)。例如：(5) + (5) = (0 -> 1)

解題：

第一版

# 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
        """
        sentinel = cur = ListNode(0)
        carry = 0
        while l1 != None or l2 != None or carry != 0:
            a = l1.val if l1 != None else 0
            b = l2.val if l2 != None else 0
            x = a + b + carry
            carry = x // 10
            x %= 10
            node = ListNode(x)
            cur.next = node
            cur = cur.next
            if l1 != None:
                l1 = l1.next
            if l2 != None:
                l2 = l2.next
        
        return sentinel.next

Runtime: 148 ms

第二版：簡(jiǎn)化代碼。

# 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
        """
        sentinel = cur = ListNode(0)
        carry = 0
        while l1 != None or l2 != None or carry != 0:
            if l1:
                carry += l1.val
                l1 = l1.next
            if l2:
                carry += l2.val
                l2 = l2.next
            
            cur.next = ListNode(carry % 10)
            carry //= 10
            cur = cur.next
        
        return sentinel.next

Runtime: 119 ms 運(yùn)行速度略微提升