Leetcode::Add Two Numbers

Problem

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 contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:

The number of nodes in each linked list is in the range [1, 100] .

0 <= Node.val <= 9

It is guaranteed that the list represents a number that does not have leading zeros.

Guess

はなぜか珍しいタイプの接続リストです.

接続リストを逆順に並べば...

クラスなので、回帰を利用してタイプを変換する必要があると思います.

Solution 1

これはリストタイプの演算方法を最大限に利用する方法である.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
    
class Solution:

    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        
        ans1, ans2 = [], []
        
            
        def dfs(head, ans):
            if head == None:
                return

            ans.append(head.val)
            dfs(head.next, ans)

        def makeListNode(prev, idx):
            if idx == len(ans):
                prev.next = None
                answer = prev
                return
                
            tmp = ListNode(ans[idx])
            prev.next = tmp
            answer = prev
            makeListNode(tmp, idx+1)

        dfs(l1, ans1)
        dfs(l2, ans2)

        ans1.reverse()
        ans2.reverse()

        ans1 = "".join(list(map(str, ans1)))
        ans2 = "".join(list(map(str, ans2)))
        
        ans = list(str(int(ans1) + int(ans2)))
        ans.reverse()
      
      	
        head = ListNode(ans[0])
        answer = head
        makeListNode(head, 1)
        
        return answer

これは.惨めである
再帰的であるため,再複文よりもスタックスペースが多く,+タイプ変換が複数回行われるのは当然の結果である.
carryを用いて初等数学加算を実現することは定解である.
サイトを手に入れたので、もう一度開けてみましょう.

Solution 1

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

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        answer = ListNode()
        head = answer
        carry = 0
        
        while l1 or l2 or carry:
            val1 = l1.val if l1 else 0
            val2 = l2.val if l2 else 0
            carry, res = divmod(val1 + val2 + carry, 10)
            
            head.next = ListNode(res)
            head = head.next
            
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
            
        return answer.next

振り返る

簡単に考えるのはかえって簡単な問題だ.

ずっと方法を使いたいと思っていたので、不要なタイプ変換や再帰演算などが増えました.

もう一度慎重に考えてから解決してもいいです.