Leetcode 2. Add Two Numbers (Medium)

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.

Solution 1 

Approach 1: Elementary Math

这里用链表结点相加计算，将l1,l2的和加到l1中。
1. 考虑链表长度：l1 < l2：将l1的尾指针指向l2 -- 将来并入l1
　　　　　　　　l1 >= l2: 继续在l1运算
2. 考虑最后一位进位：加入新的进位结点。
　　　　　　　　　　由于p1最后一定指向None而不能指向进位结点，故用tail指针标记每次的尾结点，
　　　　　　　　　　tail.next = ListNode(carry)使得l1与进位结点相连接。

 1 class Solution:
 2     def addTwoNumbers(self, l1, l2):
 3         """
 4         :type l1: ListNode
 5         :type l2: ListNode
 6         :rtype: ListNode
 7         """
 8         p1, p2 = l1, l2
 9         carry = 0
10         tail = p1
11         while p1 is not None and p2 is not None:
12             cnt = p1.val + p2. val + carry
13             p1.val = cnt % 10
14             carry = cnt // 10
15             tail = p1
16             p1 = p1.next
17             p2 = p2.next
18         
19         if p2 is not None:
20             tail.next = p2
21             tail = tail.next
22             p1 = tail
23             
24         while p1 is not None:
25             cnt = p1.val + carry
26             print(carry, p1)
27             p1.val = cnt % 10
28             carry = cnt // 10
29             tail = p1
30             p1 = p1.next
31             
32         if carry > 0:
33             tail.next = ListNode(carry)
34             tail = tail.next
35             tail.next = None
36         return l1

Beats: 95.42%
Runtime: 108ms

Approach 2. Turn into Integers

这个解法也许不是出题人希望的正规解法：
先将linked list表示的数，用整数表示，
然后整数相加，将所得和，用%/分解，表示成linked list形式。

Notice:

这里注意，所得和为0的情况不要忽略，要单独讨论。

 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         def getInteger(l):
15             res = 0
16             cnt = 1
17             while l is not None:
18                 res = res + l.val * cnt
19                 cnt *= 10
20                 l = l.next
21             return res
22         sum_integer = getInteger(l1) + getInteger(l2)
23         if sum_integer == 0:
24             return ListNode(0)
25         
26         head = ListNode(0)
27         l = head
28         while sum_integer > 0:
29             l.next = ListNode(sum_integer % 10)
30             l = l.next
31             sum_integer /= 10
32         return head.next

Beats: 94.47%
Runtime: 68ms