[LeetCode]2.两数相加（C++）

1.代码

/**
 * 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) {
        bool flag = true;
        ListNode* l3 = new ListNode;
        ListNode* point = l3;
        int carry = 0;
        while(l1 != nullptr && l2 != nullptr){
            point->val = (l1->val + l2->val + carry) % 10;
            carry = (l1->val + l2->val + carry) / 10;
            l1 = l1->next;
            l2 = l2->next;
            if (!(l1 == nullptr && l2 == nullptr)){
                point->next = new ListNode;
                point = point->next;}
            if(l2 == nullptr) flag = false;
        }
        if(flag){
            while(l2 != nullptr){
                point->val = (l2->val + carry) % 10;
                carry = (l2->val + carry) / 10;
                l2 = l2->next;
                if(l2 != nullptr){
                    point->next = new ListNode;
                    point = point->next;}
            }
        }
        else{
            while(l1 != nullptr){
                point->val = (l1->val + carry) % 10;
                carry = (l1->val + carry) / 10;
                l1 = l1->next;
                if(l1 != nullptr){
                    point->next = new ListNode;
                    point = point->next;}
            }
        }
        if(carry){
            point->next = new ListNode(carry);
        }
        return l3;

    }
};

2.思路

先对两链表都非空的部分做加法，再分别对l1较长，l2较长和l1,l2一样长的情况分别讨论。最后一位如果有进位就在l3末尾加上最高位，如果没有进位则无需额外处理。