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
解题思路:很简单,主要考察链表的操作,对链表的操作还不是很熟,另外没有见过这样的结构体,以后还要再做一遍。
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; 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; } };