leetcode algorithm2 add two number

原题链接

问题

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

思路

没啥好思路的，就当复习写代码了。处理好进位即可。注意他给的数据结构。还有就是别忘了判断None值。

代码

python实现

# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        copy = None        # two non-empty list        if (l1 != None) and (l2 != None):            # if there is still one list that have digits            reply = None            carry = 0            while(l1 !=None or l2 != None):                if (l1 != None):                    value1 = l1.val                else:                    value1 = 0                if (l2 != None):                    value2 = l2.val                else:                    value2 = 0                sum = value1 + value2 + carry                digit = sum % 10                carry = sum // 10                if reply == None:                    reply = ListNode(digit)                    copy = reply                else:                    reply.next = ListNode(digit)                    reply = reply.next                if(l1 != None):                    l1 = l1.next                if(l2 != None):                    l2 = l2.next            # in the end, there might be a carry            if carry != 0:                reply.next = ListNode(carry)            # copy equal to reply?            return copy    def addTwoNumbersBetter(self, l1, l2):        carry = 0        root = n = ListNode(0)        while l1 or l2 or carry:            v1 = v2 = 0            if l1:                v1 = l1.val                l1 = l1.next            if l2:                v2 = l2.val                l2 = l2.next            carry, val = divmod(v1+v2+carry,10)            n.next = ListNode(val)            n = n.next        return root.nextinstance = Solution()list1 = ListNode(2)list1.next = ListNode(4)list1.next.next = ListNode(3)list2 = ListNode(5)list2.next = ListNode(6)list2.next.next = ListNode(4)#reply = instance.addTwoNumbers(list1,list2)reply = instance.addTwoNumbersBetter(list1,list2)while(reply != None):    print(reply.val)    reply = reply.next