LeetCode 2. Add Two Numbers

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

把较短的字符串的数字加到较长的字符串上，唯一注意的是进位问题

c++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 *p = l1;
        ListNode *q = l2;
        
        int len1 = 0;
        int len2 = 0;
        
        while(p != nullptr)
        {
            ++ len1;
            p = p -> next;
        }
        
        while(q != nullptr)
        {
            ++ len2;
            q = q -> next;
        }
        
        ListNode *res = nullptr;
        
        if(len1 >= len2)
        {
            res = l1;
            p = l1;
            q = l2;
        }
        else 
        {
            res = l2;
            p = l2;
            q = l1;
        }
        
        while(q != nullptr)
        {
            p -> val += q -> val;
            p = p -> next;
            q = q -> next;
        }
        
        ListNode *r = res;
        
        while(r -> next != nullptr)
        {
            if(r -> val >= 10)
            {
                int t = r -> val / 10;
                r -> val = r -> val % 10;
                r -> next -> val += t;
            }
                
            r = r -> next;
        }
        
        if(r -> val >= 10)
        {
            int t = r -> val / 10;
            r -> val %= 10;
            ListNode *tmp = new ListNode(t);
            r -> next = tmp;
        }
        return res;
    }
};