Description
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.
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8
Discussion
这个问题比较简单，相当于实现一个基于链表的超大数加法。需要注意的地方就是进位的处理。
算法的时间复杂度为O(n)。
C++ Code
/**
* 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 * answer = new ListNode(0);
ListNode * p = answer;
int flag = 0;
while(l1 || l2 || flag)
{
int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + flag;
p->next = new ListNode(sum % 10);
p = p->next;
flag = sum / 10;
l1 = l1 ? l1->next : l1;
l2 = l2 ? l2->next : l2;
}
return answer->next;
}
};