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
思路非常简单，利用两个指针分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。
class Solution {public:ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {// Start typing your C/C++ solution below// DO NOT write int main() function//ListNode *pResult = NULL;//ListNode **pCur = &pResult;ListNode rootNode(0);ListNode *pCurNode = &rootNode;int a = 0;while (l1 || l2){int v1 = (l1 ? l1->val : 0);int v2 = (l2 ? l2->val : 0);int temp = v1 + v2 + a;a = temp / 10;ListNode *pNode = new ListNode((temp % 10));pCurNode->next = pNode;pCurNode = pNode;if (l1)l1 = l1->next;if (l2)l2 = l2->next;}if (a > 0){ListNode *pNode = new ListNode(a);pCurNode->next = pNode;}return rootNode.next;}};
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一起吃早餐，午餐，晚餐。或许吃得不好，
