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

思路

• 很简单的问题，主要是注意进位的问题，以及最后特殊情况，即最后有个进位

代码

• list1和list2的长度分别为n,m。算法复杂度为O(max(n, m))

/*
 * 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 *res = NULL, *ptr = NULL;
        int count = 0, sum = 0;
        
        while(l1 || l2){
            sum = 0;
            if(l1){
                sum += l1->val;
                l1 = l1->next;
            }
            if(l2){
                sum += l2->val;
                l2 = l2->next;
            }
            
            sum += count;
            count = sum / 10;
            sum %= 10;
            
            ListNode *tmp = new ListNode(sum);
            if(!res){
                res = tmp;
                ptr = tmp;
            }
            else{
                ptr->next = tmp;
                ptr = ptr->next;
            }
        }
        
        //特殊情况，比如9 + 9 = 18，最后的 1  
        if(count > 0){
            ListNode *tmp = new ListNode(count);
            ptr->next = tmp;
        }
        
        return res;
    }
};