5-4打卡,力扣2. 两数相加

2. 两数相加

难度中等

给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。

请你将两个数相加,并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外,这两个数都不会以 0 开头。

 

示例 1:

输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807.

示例 2:

输入:l1 = [0], l2 = [0]
输出:[0]

示例 3:

输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1]

 

提示:

  • 每个链表中的节点数在范围 [1, 100] 内
  • 0 <= Node.val <= 9
  • 题目数据保证列表表示的数字不含前导零
    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     struct ListNode *next;
     * };
     */
    
    
    struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
        struct ListNode *dummyHead = (struct ListNode *)malloc(sizeof(struct ListNode));
        dummyHead->val = 0;
        dummyHead->next = NULL;
        struct 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 = (struct ListNode *)malloc(sizeof(struct ListNode));
            curr->next->val = sum % 10;
            curr->next->next = NULL;
            curr = curr->next;
            if (p != NULL) p = p->next;
            if (q != NULL) q = q->next;
        }
        if (carry > 0) {
            curr->next = (struct ListNode *)malloc(sizeof(struct ListNode));
            curr->next->val = carry;
            curr->next->next = NULL;
        }
        return dummyHead->next;
    }

     

posted @ 2023-05-04 22:51  aallofitisst  阅读(13)  评论(0)    收藏  举报