[LeetCode] Add Two Numbers 链表

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

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 Linked List Math

 

---

　　这题是链表遍历问题，容易处理。

 

#include <iostream>
using namespace std;
/**
 * 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) {
        if(l1==NULL)    return l2;
        if(l2==NULL)    return l1;
        int addOne=0;
        ListNode ret(0);
        ListNode *p1=l1,*p2=l2,*pret=&ret;
        while(1){
            if(p1==NULL&&p2==NULL&&addOne==0)   break;
            if(p1==NULL&&p2==NULL){
                pret->next = new ListNode((addOne)%10);
                pret = pret->next;
                addOne = 0;
                continue;
            }
            if(p1==NULL){
                pret->next = new ListNode((p2->val + addOne)%10);
                pret = pret->next;
                addOne = (p2->val + addOne)/10;
                p2=p2->next;
                continue;
            }
            if(p2==NULL){
                pret->next = new ListNode((p1->val + addOne)%10);
                pret = pret->next;
                addOne = (p1->val + addOne)/10;
                p1=p1->next;
                continue;
            }
            pret->next = new ListNode((p1->val + p2->val + addOne)%10);
            pret = pret->next;
            addOne = (p1->val + p2->val + addOne)/10;
            p1=p1->next;
            p2=p2->next;
        }
        return ret.next;
    }
};

int main()
{

    return 0;
}