2 .Two numbers

2024-01-01

字数统计: 388字 | 阅读时长: 2min

topic：

Thought：

I originally wrote the stupidest way，把两个Linked转换成数字，Then add，再转换成Linked。But it’s strange that I was right when I was running locally，But it’s wrong when you submit，
What is deducted is a very strange oneprecompiled.listnode.ListNode，But mine is'__main__.ListNode'。
So I can only watch0x3fs answer。

Two node values each time`l1.val`，`l2.val`In -digit`carry`Add，Divide 10 The remaining number is the digital digit that the current node needs to be saved，Divide10The business is the new position value

Code implementation，There is a trick to simplify code：ifrecursion中发现l2Length ratio ratiol1Longer，So it can be exchangedl1and l2，ensure l1Not an empty node，So as to simplify code logic。

Code：

class Solution:
    # l1 and l2 Node for current traversal，carry Be in place
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode], carry=0) -> Optional[ListNode]:
        if l1 is None and l2 is None:  # recursion边界：l1 and l2 All empty nodes
            return ListNode(carry) if carry else None  # If you are in place，Just create an additional node
        if l1 is None:  # if l1 Empty，Then then l2 一定Not an empty node
            l1, l2 = l2, l1  # exchange l1 and l2，ensure l1 non empty，从而简化Code
        carry += l1.val + (l2.val if l2 else 0)  # 节点值andcarry加在一起
        l1.val = carry % 10  # Each node saves a digital position
        l1.next = self.addTwoNumbers(l1.next, l2.next if l2 else None, carry // 10)  # carry
        return l1

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode], carry = 0) -> Optional[ListNode]:
        l1_list = []
        l2_list=[]
        def recursion_list(node:ListNode, listt):
            if not node:
                return listt
            listt.append(node.val)
            return recursion_list(node.next, listt)
        l1_list = recursion_list(l1, l1_list)
        l2_list = recursion_list(l2, l2_list)
        l1_list = ''.join(map(str, l1_list))
        l2_list = ''.join(map(str, l2_list))
        l3 = str(int(l1_list)+int(l2_list))
        print(l3)
        def recursion_linkArray(num):
            if not num:
                return None
            head = ListNode(0)
            current = head
            for c in num:
                current.next = ListNode(int(c))
                current = current.next
            return head.next
        head = recursion_linkArray(l3[::-1])
        print(recursion_list(head, []))
        print(type(l1), type(head))
        return head