Leetcode|短链表后补零|2. 两数相加

/**
 * 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) {
        auto prehead = new ListNode();
        auto pre = prehead;
        // 进位
        int plus = 0;
        while (l1 || l2) {
            int sum = 0;  // 同1位下l1和l2的加和
            if (l1) {     // 若当前l1不为空，则sum加上当前l1节点的值
                sum += l1->val;
                l1 = l1->next;
            }
            if (l2) {
                sum += l2->val;
                l2 = l2->next;
            }
            // 若l1和l2节点+上1位进位加和>10
            if (sum + plus >= 10) {
                pre->next = new ListNode(sum + plus - 10);
                plus = 1;
            } else {
                pre->next = new ListNode(sum + plus);
                plus = 0;
            }
            pre = pre->next;
        }
        // 考虑最后1位加和可能产生的进位
        if (plus == 1) pre->next = new ListNode(1);
        return prehead->next;
    }
};