Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    int  listLen(ListNode* head)
    {
        int res = 0;
        while(head!=NULL){
            res++;
            head = head->next;
        }
        return res;
    }
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int len1 = listLen(l1);
        int len2 = listLen(l2);
        
        ListNode* head1 = len1>len2 ? l1:l2;
        ListNode* head2 = len1>len2 ? l2:l1;
        ListNode* head = head1;
        int c = 0;
        ListNode* pre = NULL;
        while(head1!=NULL && head2!=0){
            int tmp = (head1->val + head2->val + c )%10;
            c = (head1->val + head2->val + c )/10;
            head1->val = tmp;
            pre = head1;
            head1 = head1->next;
            head2 = head2->next;
            cout<<"tmp = "<<tmp<<",c="<<c<<endl;
        }
        while(head1!=NULL){
           
            int tmp = (head1->val + c)%10;
            c = (head1->val + c)/10;
            head1->val = tmp;
            pre = head1;
            head1 = head1->next;
             cout<<"tmp = "<<tmp<<",c="<<c<<endl;
        }
        
        while(head2!=NULL){
            
            int tmp = (head2->val + c)%10;
            c = (head2->val + c)/10;
            head2->val = tmp;
            pre = head2;
            head2 = head2->next;
            cout<<"tmp = "<<tmp<<",c="<<c<<endl;
        }
        
        if(c!=0){
            ListNode *node = new ListNode(c);
            pre->next = node;
        }
        return head;
    }
};
原文地址:https://www.cnblogs.com/zengzy/p/5002079.html