002.md

2. Add Two Numbers

难度: Medium

刷题内容

原题连接

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

内容描述

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.

解题方案

思路 1 - 时间复杂度: O(N) - 空间复杂度: O(N)

全部变成数字做加法再换回去呗，这多暴力，爽！

beats 54.35%

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1:
            return l2
        if not l2:
            return l1
        
        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]])
        
        tmp = str(int(num1) + int(num2))[::-1]
        res = ListNode(int(tmp[0]))
        run_res = res
        for i in range(1, len(tmp)):
            run_res.next = ListNode(int(tmp[i]))
            run_res = run_res.next
        return res

思路 2 - 时间复杂度: O(N) - 空间复杂度: O(1)

可以使用递归，每次算一位的相加, beats 83.40%

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