题目描述
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
这道并不是什么难题,算法很简单,链表的数据类型也不难。就是建立一个新链表,然后把输入的两个链表从头往后撸,每两个相加,添加一个新节点到新链表后面,就是要处理下进位问题。还有就是最高位的进位问题要最后特殊处理一下。代码如下:
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode *res = new ListNode(-1);
ListNode *cur = res;
int carry = 0;
while (l1 || l2) {
int n1 = l1 ? l1->val : 0;
int n2 = l2 ? l2->val : 0;
int sum = n1 + n2 + carry;
carry = sum / 10;
cur->next = new ListNode(sum % 10);
cur = cur->next;
if (l1) l1 = l1->next;
if (l2) l2 = l2->next;
}
if (carry) cur->next = new ListNode(1);
return res->next;
}
};