Leetcode [Add Two Numbers]

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本文介绍了一种使用链表表示数字并实现两个链表所代表数字相加的方法。具体思路是遍历两个链表，逐位相加并考虑进位情况，最终返回新的链表作为两数之和。

Problem：Add Two Numbers

Question

You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

思路

用链表表示一个数，求两个数的和
给出的实例是 342 + 465 = 807（链表是反序的）
所以基本思路是，两个链表对应位置相加，从head加到tail，进位记为carry.

代码如下：

/**
 * 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, val = 0;
        ListNode* result = new ListNode(0);
        ListNode* head = result;
        result->val = (l1->val + l2->val) % 10;
        carry = (l1->val+l2->val) / 10;
        while(l1->next != NULL || l2->next != NULL) {
            if (l1->next != NULL && l2->next != NULL) {
                l1 = l1->next;
                l2 = l2->next;
                val = l1->val+l2->val+carry;
            }
            else if (l1->next == NULL) {
                l2 = l2->next;
                val = l2->val+carry;
            }
            else if (l2->next == NULL) {
                l1 = l1->next;
                val = l1->val+carry;
            }
            result->next = new ListNode(val%10);
            result = result->next;
            carry = val/10;
        }
        if(carry != 0) {
            result->next = new ListNode(carry);
            result = result->next;   
        }
        return head;
    }
};