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：

本题，考察了按位加法与链表的构建，思路很简单直接，按位相加，判断有无进位，再构建链表

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 sum,carry = 0;        ListNode* temp = new ListNode(0);        ListNode* head = temp;        while(l1||l2)        {            if(l1!=NULL&&l2!=NULL)            {                sum = l1->val+l2->val+carry;//按位相加                carry = sum / 10;                ListNode* l = new ListNode(sum%10);                temp->next = l ;//构建链表（后向链表）                temp = l;                l1 = l1->next;                l2 = l2->next;            }                        else if(!l1&&l2!=NULL)            {                sum = l2->val+carry;                carry = sum / 10;                ListNode* l = new ListNode(sum%10);                temp->next = l ;                temp = l;                  l2 = l2->next;            }            else if(!l2&&l1!=NULL)            {                sum = l1->val+carry;                carry = sum / 10;                ListNode* l = new ListNode(sum%10);                temp->next = l ;                temp = l;                  l1 = l1->next;            }                     }        if(carry!=0)        {            ListNode* l = new ListNode(carry);            temp->next = l ;            temp = l;         }        temp->next = NULL;        head = head->next;        return head;    }};