题目

2. 两数相加

答案

链表

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/**
 * 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) {
        // 链表定义头指针和尾指针
        ListNode *head = nullptr, *tail = nullptr;
        //进制位
        int carry_bit = 0;
        while (l1 || l2) 
        {
            int number1 = 0, number2 = 0, sum = 0;
            //获取该进制位数的值
            if (l1) number1 = l1->val;
            if (l2) number2 = l2->val;
            //同位相加
            sum = number1 +number2+ carry_bit;
            //初始第一次
            if (!head) 
                head = tail = new ListNode(sum % 10);
            else 
            {   
                //在堆上新建内存，分配给尾指针的下一结点
                tail->next = new ListNode(sum % 10);
                //将尾指针设置新建的结点
                tail = tail->next;
            }
            //进制位
            carry_bit = sum/10;
            //头指针向后移
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        }
        //如果最后一位进一的话，就将尾指针设置成1
        if (carry_bit==1)
        {
            tail->next = new ListNode(1);
            tail = tail->next;
        }
        //返回链表的头指针即可
        return head;
    }
};

__END__