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

Solution：
Keep track of the carry using a variable and simulate digits-by-digits sum starting from the head of list, which contains the least-significant digit.
下图模拟了计算的过程：有点像在大数运算中将数倒着存的方法，

Figure 1. Visualization of the addition of two numbers:
结果为：342+465=807.
Each node contains a single digit and the digits are stored in reverse order.
还要考虑以下几种特殊情况：

原理知道了，代码如下：

#include <iostream>using namespace std;struct ListNode{        int val;        ListNode *next;        ListNode(int x) : val(x), next(NULL) {}};ListNode* initList(int arr[],int n){        ListNode *head = NULL,*cur = NULL;        for(int i = 0; i < n; ++i)        {                ListNode* nodes = new ListNode(arr[i]);                if(head == NULL)                        head = nodes;                if(cur != NULL)                        cur->next = nodes;                cur = nodes;        }        return head;}ListNode* addSum(int num1,int num2,int &sign){    int num = num1 + num2 + sign;    if(num >= 10)    {        num -= 10;        sign = 1;    }    else    {        sign = 0;    }    ListNode *nodes = new ListNode(num);    return nodes;}ListNode* addTwoNumbers(ListNode* l1, ListNode* l2){        if(l1 == NULL || l2 == NULL)                return NULL;        ListNode* list1 = l1,*list2 = l2;        ListNode* head = NULL,*cur = NULL,*last = NULL;        int sign = 0,num;        //sign是在模拟手算加法的时候，表示是否进位的标志。        while(list1 != NULL && list2 != NULL)        {                ListNode* nodes = addSum(list1->val,list2->val,sign);                if(head == NULL)                        head = nodes;                if(cur != NULL)                        cur->next = nodes;                cur = nodes;                list1 = list1->next;                list2 = list2->next;        }        while(list1 != NULL)        {                ListNode* nodes = addSum(list1->val,0,sign);                if(cur != NULL)                        cur->next = nodes;                cur = nodes;                list1 = list1->next;        }                while(list2 != NULL)        {                ListNode* nodes = addSum(0,list2->val,sign);                if(cur != NULL)                        cur->next = nodes;                cur = nodes;                list2 = list2->next;        }        if(sign)        {                ListNode* nodes = new ListNode(sign);                cur->next= nodes;        }        return head;}int main(){        int arr1[] = {2,4,3};        int arr2[] = {5,6,4};        ListNode *l1 = initList(arr1,sizeof(arr1)/sizeof(arr1[0]));        ListNode *l2 = initList(arr2,sizeof(arr2)/sizeof(arr2[0]));        ListNode *l3 = addTwoNumbers1(l1,l2);        return 0;}

写完发现，虽然代码提交成功，但是发现在讨论中有一种更加简洁高效的方法。代码如下：

ListNode* addTwoNumbers1(ListNode *l1,ListNode *l2){    ListNode *list1 = l1;    ListNode *list2 = l2;    ListNode *head = new ListNode(0);    ListNode *d = head;    int sum = 0;    while(list1 != NULL || list2 != NULL)    {        sum /= 10;        if(list1 != NULL)        {            sum += list1->val;            list1 = list1->next;        }        if(list2 != NULL)        {            sum += list2->val;            list2 = list2->next;        }        d->next = new ListNode(sum % 10);        d = d->next;    }    if(sum / 10 == 1)        d->next = new ListNode(1);    return head->next;}