tiktokOA Add Two Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/HuiFeiDeTuoNiaoGZ/article/details/120591369

该博客介绍了一个C++实现的解决方案，用于将两个单链表表示的非负整数相加。通过遍历两个链表并处理进位，最终合并成一个新的链表表示结果。代码中定义了链表节点结构体，并提供了详细的加法计算逻辑。

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自己做的

/**
 * 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* start = l1;
        int over = 0;
        while(l1->next != nullptr && l2->next != nullptr){
            int sum = over + l1->val + l2->val;
            if(sum >= 10){
                over = 1;
                sum -= 10;
            }
            else{
                over = 0;
            }
            l1->val = sum;
            l1 = l1->next;
            l2 = l2->next;
        }
        int sum = over + l1->val + l2->val;
        if(sum >= 10){
            over = 1;
            sum -= 10;
        }
        else{
            over = 0;
        }
        l1->val = sum;
        
        if(!l1->next){
            l1->next = l2->next;
        }
        while(over){
            if(!l1->next){
                l1->next = new ListNode(over, nullptr);
                return start;
            }
            else{
                sum = over + l1->next->val;
                    if(sum >= 10){
                    over = 1;
                    sum -= 10;
                }
                else{
                    over = 0;
                }
                l1->next->val = sum;
                l1 = l1->next;
            }
        }
        return start;
    }
};

答案简洁一点

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}