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

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。

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        p1, p2 = l1, l2
        carry = 0
        tail = p1
        while p1 is not None and p2 is not None:
            cnt = p1.val + p2. val + carry
            p1.val = cnt % 10
            carry = cnt // 10
            tail = p1
            p1 = p1.next
            p2 = p2.next
        
        if p2 is not None:
            tail.next = p2
            tail = tail.next
            p1 = tail
            
        while p1 is not None:
            cnt = p1.val + carry
            print(carry, p1)
            p1.val = cnt % 10
            carry = cnt // 10
            tail = p1
            p1 = p1.next
            
        if carry > 0:
            tail.next = ListNode(carry)
            tail = tail.next
            tail.next = None
        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。

# 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
        """
        def getInteger(l):
            res = 0
            cnt = 1
            while l is not None:
                res = res + l.val * cnt
                cnt *= 10
                l = l.next
            return res
        sum_integer = getInteger(l1) + getInteger(l2)
        if sum_integer == 0:
            return ListNode(0)
        
        head = ListNode(0)
        l = head
        while sum_integer > 0:
            l.next = ListNode(sum_integer % 10)
            l = l.next
            sum_integer /= 10
        return head.next

Beats: 94.47%
Runtime: 68ms