试题

试题链接

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

试题大意

把两个整数分别用两个反序链表的形式来表示（十进制表示法的从低位到高位依次存入链表）。对这两个整数求和。结果依然以反序链表表示。

解题思路

这是纯粹的十进制大数加法模拟题。

循环遍历链表。对于每轮循环，记录第一个链表（如果未遍历到末尾）与第二个链表（如果未遍历到末尾）与上次进位之和。把结果的个位数放在新链表末尾，记录进位，供下次使用。最后一次要把进位（如果不是0）插入新链表末尾。

源代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int carry = 0, current_digit = 0;
        ListNode* result = NULL;
        while (l1 != NULL || l2 != NULL)
        {
            current_digit = carry;
            if (l1 != NULL)
            {
                current_digit += l1->val;
                l1 = l1->next;
            }

            if (l2 != NULL)
            {
                current_digit += l2->val;
                l2 = l2->next;
            }

            carry = current_digit / 10;
            current_digit %= 10;

            ListNode *temp = new ListNode(current_digit);
            appendNode(result, (new ListNode(current_digit)));
        }
        if (carry != 0)
            appendNode(result, (new ListNode(carry)));
        return result;
    }
private:
    void appendNode(ListNode *&l, ListNode *e)
    {
        if (l == NULL)
        {
            l = e;
            return;
        }

        ListNode *q = l;
        ListNode *p = q->next;
        while (p != NULL)
        {
            q = p;
            p = q->next;
        }
        q->next = e;
    }
};