leetcode 2 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

我的想法有点复杂，就是两个加数所有逆置，就是遍历放到栈里面，完后出栈加起来放在容器中。返回链表

可是这个算法好像有错误。

很不解：

// addtwonumber.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <stack>
#include <vector>

using namespace std;

struct ListNode {
	     int val;
	     ListNode *next;
	     ListNode(int x) : val(x), next(NULL) {}
	 };

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
{
	ListNode* temp1 = l1;
	ListNode* temp2 = l2;
	stack<int> s_int1,s_int2;
	vector<int> v_int;


	
	int length = 0,length1 = 0,length2 =0;
	while(temp1)
	{
		++length1;
		
		s_int1.push(temp1->val);
		temp1 = temp1->next;
		
	}

	while(temp2)
	{
		++length2;
		

		s_int2.push(temp2->val);
		temp2 = temp2->next;
	}

	length = length1>length2?length1:length2;
	int jinwei = 0;
	int flag = 0;
	int value1 = 0;
	int value2 = 0;

	while(!s_int1.empty()||!s_int2.empty()||flag)
	{
		if (!s_int1.empty())
		{
			value1 = s_int1.top();
			s_int1.pop();
		}
		else
		{
		value1 = 0;
		}
		if (!s_int2.empty())
		{

			value2 = s_int2.top();
			s_int2.pop();
		}
		else
		{
		value2 = 0;
		}
		
		

		jinwei = value1 + value2  + flag;
		
			v_int.push_back(jinwei%10);
			jinwei = jinwei/10;
			

	}



	ListNode* result = (ListNode* )malloc(sizeof(ListNode));
	result->val = v_int[0];
	
	ListNode* t = NULL;
	
	ListNode* r = result;
	for(int i = 1; i <length;++i)
	{
		t = (ListNode* )malloc(sizeof(ListNode));
		t->val = v_int[i];
		t->next = NULL;
		r->next = t;
		r = t;
	   

	}
	return result;

}

int _tmain(int argc, _TCHAR* argv[])
{

	ListNode n1(1),n2(2),n3(3); 

	n1.next = &n2;
	n2.next = &n3;

	ListNode n4(9),n5(5),n6(6);
	n4.next = &n5;
	//n5.next = &n6;


	ListNode l1(1),l2(0);

	addTwoNumbers(&n1,&n4);
	addTwoNumbers(&l1,&l2);
	return 0;
}

正确的代码：很easy介绍

基本思路差点儿相同

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) 
{
    ListNode preHead(0), *p = &preHead;
    int extra = 0;
    while (l1 || l2 || extra) {
        if (l1) extra += l1->val, l1 = l1->next;
        if (l2) extra += l2->val, l2 = l2->next;
        p->next = new ListNode(extra % 10);
        extra /= 10;
        p = p->next;
    }
    return preHead.next;
}

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
 {
    ListNode preHead(0), *p = &preHead;
    int extra = 0;
    while (l1 || l2 || extra) {
        int sum = (l1 ?
 l1->val : 0) + (l2 ? l2->val : 0) + extra;
        extra = sum / 10;
        p->next = new ListNode(sum % 10);
        p = p->next;
        l1 = l1 ?
 l1->next : l1;
        l2 = l2 ? l2->next : l2;
    }
    return preHead.next;
}

python代码：

class Solution:
# @return a ListNode
def addTwoNumbers(self, l1, l2):
    carry = 0
    root = n = ListNode(0)
    while l1 or l2 or carry:
        v1 = v2 = 0
        if l1:
            v1 = l1.val
            l1 = l1.next
        if l2:
            v2 = l2.val
            l2 = l2.next
        carry, val = divmod(v1+v2+carry, 10)
        n.next = ListNode(val)
        n = n.next
    return root.next


def addTwoNumbers(self, l1, l2):
        carry = 0;
        res = n = ListNode(0);
        while l1 or l2 or carry:
            if l1:
                carry += l1.val
                l1 = l1.next;
            if l2:
                carry += l2.val;
                l2 = l2.next;
            carry, val = divmod(carry, 10)
            n.next = n = ListNode(val);
        return res.next;