LeetCode2——Add Two Numbers

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题目链接

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
Explanation: 342 + 465 = 807.

代码

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

class Solution:
    def addTwoNumbers(self, l1: 'ListNode', l2: 'ListNode') -> 'ListNode':
        if l1 == None:
            return l2
        if l2 == None:
            return l1

        returnlist = ListNode(0)
        head = returnlist
        point1 = l1
        point2 = l2
        carry = 0
        while point1 != None and point2 != None:
            if point1.val + point2.val + carry >= 10:
                returnlist.val = (point1.val + point2.val + carry) % 10
                carry = 1
            else:
                returnlist.val = (point1.val + point2.val + carry)
                carry = 0

            if carry == 0 and point1.next == None and point2.next == None:
                return head
            else:
                newNode = ListNode(0)
                returnlist.next = newNode

            if point1 != None:
                point1 = point1.next
            if point2 != None:
                point2 = point2.next
            returnlist = returnlist.next

        while point1 != None:
            if carry + point1.val >= 10:
                returnlist.val = (carry + point1.val) % 10
                carry = 1
            else:
                returnlist.val = carry + point1.val
                carry = 0
            point1 = point1.next
            if point1 != None or carry == 1:
                returnlist.next = ListNode(0)
                returnlist = returnlist.next

        while point2 != None:
            if carry + point2.val >= 10:
                returnlist.val = (carry + point2.val) % 10
                carry = 1
            else:
                returnlist.val = carry + point2.val
                carry = 0
            point2 = point2.next
            if point2 !=None or carry == 1:
                returnlist.next = ListNode(0)
                returnlist = returnlist.next

        if carry == 1 and point1 == None and point2 == None:
            returnlist.val += 1
            return head
        return head