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); 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); 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); 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; } };
