2. Add Two Numbers

题目

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

第一种：恢复输入的两个数，然后相加，再转换回去。这种方法会遇到数值范围的问题，不予考虑。
第二种：由于数字是以个位开头的，可以直接相加进位到下一位，最后如果多出进位的话再加一位即可。

实现

/**
 * 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 ones = 0;
        int tens = 0;
        int rlt = 0;
        ListNode* ans;
        ListNode* last = NULL;
        
        while(l1!=NULL || l2!=NULL){
            if(l1 == NULL){
                rlt = l2->val + tens;
                l2 = l2->next;
            }
            else if(l2 == NULL){
                rlt = l1->val + tens;
                l1 = l1->next;
            }
            else{
                rlt = l1->val + l2->val + tens;
                l1 = l1->next;
                l2 = l2->next;
            }
            ones = rlt % 10;
            tens = rlt / 10;
            ListNode* node = new ListNode(ones);
            if(last == NULL){
                ans = node;
                last = node;
            }
            else{
                last->next = node;
                last = node;
            }
        }
        if(tens>0){
            ListNode* node = new ListNode(tens);
            last->next = node;
        }
        return ans;
    }
};

主要考察对于链表的理解，不多说了。