两数之和解法

/**
 * 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* out = new ListNode;
        ListNode* L = new ListNode;
        L = out;

        int cin = 0;

        while(l1 != nullptr || l2 != nullptr){
            out->val = l1->val + l2->val + cin;
            if(out->val>=10){
                out->val %= 10;
                cin = 1;
            }else cin = 0;
            if(l1->next == nullptr && l2->next == nullptr){
                if(cin == 1){
                    out->next = new ListNode;
                    out = out->next;
                    out->val = 1;
                }
                out->next = nullptr;
                break;
            }
            
            out->next = new ListNode;
            out = out->next;

            if(l1->next != nullptr) l1 = l1->next;
            else l1->val = 0;
            if(l2->next != nullptr) l2 = l2->next;
            else l2->val = 0;
        }
        return L;
    }
};

两数之和解法

原文：https://www.cnblogs.com/yoshinb/p/15027568.html