力扣刷题-热题100题-第28题（c++、python）

2. 两数相加 - 力扣（LeetCode）https://leetcode.cn/problems/add-two-numbers/description/?envType=study-plan-v2&envId=top-100-liked

常规法

根据加法的规则，设置一个记位数，初始为0，遍历两个链表，相同位数相加并加上记位数得到最终的值，以个位数作为当前位数的和，十位数更新记位数。

//c++
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) 
    {
        ListNode *head=nullptr,*tail=nullptr;
        int carry=0;
        while(l1||l2)
        {
            int n1=l1?l1->val:0;
            int n2=l2?l2->val:0;
            carry=n1+n2+carry;
            if(!head)
            {
                head=tail=new ListNode(carry%10);
            }
            else
            {
                tail->next=new ListNode(carry%10);
                tail=tail->next;
            }
            carry=carry/10;
            if(l1) l1=l1->next;
            if(l2) l2=l2->next;
        }
        if(carry>0) tail->next=new ListNode(carry);
        return head;
    }
};

#python
# 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: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        if l1==None or l2==None:
            return None
        ans=ListNode()
        a=ans
        aa=0
        while l1 or l2:
            a1=l1.val if l1 else 0
            a2=l2.val if l2 else 0
            aa=aa+a1+a2
            b=aa%10
            aa=(aa//10)%10
            c=ListNode(b)
            a.next=c
            a=a.next
            if l1:
                l1=l1.next
            if l2:
                l2=l2.next
        if aa:
            c=ListNode(aa)
            a.next=c
        return ans.next