问题
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.
输入
(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出
7 -> 0 -> 8
分析
两个链表相加，生成一个和链表。同步遍历两个链表，算出当前两个节点的和sum，把sum%10插入到和链表中，把sum/10作为carry加入到下一步的sum计算。
要点
简单的按位相加，要考虑到链表的结尾以及carry的处理。
时间复杂度
O(max(m,n))，m和n分别为输入两个链表的长度。
空间复杂度
O(1)
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) {
ListNode *l3 = new ListNode(-1);
ListNode *pl3 = l3;
int carry = 0;
while (l1 || l2 || carry) {
int sum = carry;
if (l1 != NULL) {
sum += l1->val;
l1 = l1->next;
}
if (l2 != NULL) {
sum += l2->val;
l2 = l2->next;
}
carry = sum / 10;
ListNode *tmp = new ListNode(sum % 10);
pl3->next = tmp;
pl3 = tmp;
}
return l3->next;
}
};
