LeetCode - 002

咸鱼学Python

发布于 2019-06-03 16:19:02

4010

原题

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

题干

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.

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解

class Solution:    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """             val1, val2 = [l1.val], [l2.val]
        while l1.next:            val1.append(l1.next.val)            l1 = l1.next        while l2.next:            val2.append(l2.next.val)            l2 = l2.next

        num1 = ''.join([str(i) for i in val1[::-1]])        num2 = ''.join([str(i) for i in val2[::-1]])
        sums = int(num1) + int(num2)
        sums = str(sums)[::-1]
        dummy = head = ListNode(int(sums[0]))        for i in range(1, len(sums)):            head.next = ListNode(int(sums[i]))            head = head.next        return dummy

解

class Solution:    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """     
        if not l1 and not l2:            return        elif not (l1 and l2):            return l1 or l2        else:            if l1.val + l2.val < 10:                l3 = ListNode(l1.val+l2.val)                l3.next = self.addTwoNumbers(l1.next, l2.next)            else:                l3 = ListNode(l1.val+l2.val-10)
                l3.next = self.addTwoNumbers(l1.next, self.addTwoNumbers(l2.next, ListNode(1)))        return l3

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原始发表：2019-05-13，如有侵权请联系 cloudcommunity@tencent.com 删除

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