[LeetCode]2. Add Two Numbers

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;    }};