【LeetCode】2. Add Two Numbers 解题报告（Python）

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本文链接： https://blog.csdn.net/ttinch/article/details/102523986

题目地址：https://leetcode.com/problems/add-two-numbers/

题目描述

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.

解法一：先求和，再构建链表

# 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:
        num1 = ''
        num2 = ''
        while l1:
            num1 = num1+str(l1.val)
            l1 = l1.next
        while l2:
            num2 = num2+str(l2.val)
            l2 = l2.next
        add = str(int(num1[::-1])+int(num2[::-1]))[::-1]
        l = ListNode(add[0])
        tmp = l
        for i in range(1, len(add)):
            tmp.next = ListNode(add[i])
            tmp = tmp.next
        return l

解法二：采用一个进位carry方便的完成一次遍历得出结果

两个要注意的地方：如果列表长度不相等；如果列表相加完成最后仍有进位位。

# 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:
        ans = ListNode(0)
        tmp = ans
        carry = 0
        while l1 and l2:
            add = l1.val + l2.val + carry
            tmp.next = ListNode(add % 10)
            tmp = tmp.next
            carry = int(add/10)
            l1 = l1.next
            l2 = l2.next
        l = l1 if l1 else l2
        while l:
            add = l.val + carry
            tmp.next = ListNode(add % 10)
            tmp = tmp.next
            carry = int(add / 10)
            l = l.next
        if carry:
            tmp.next = ListNode(carry)
        return ans.next

解法三：递归

# 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 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