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删除前面的linklist，使用node来表示链表// You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
//
// EXAMPLE
//
//Input: (3 -> 1 -> 5), (5 -> 9 -> 2)
//
//Output: 8 -> 0 -> 8#include <iostream>
using namespace std;
struct node
{int data;node *next;
};void init(node *p,int a[], int size)
{if (a==NULL || size<0){return;}for (int i = 0; i < size; i++){node *t = new node;t->data = a[i];p->next = t;p = p->next;}p->next = NULL;
}void print(node *head)
{if (!head){return;}node *p = head->next;while (p){cout<<p->data<<" ";p = p->next;}cout<<endl;
}node * plus(node *a, node *b)
{node *pa = a->next;node *pb = b->next;node *pc = new node;node *pi = pc;int i = 0;int step = 0;while(pa!=NULL && pb!= NULL){node *t = new node;int temp = pa->data + pb->data +step;step = 0;if (temp >=10){temp %= 10;step = 1;}t->data =temp;pi->next = t;pi = t;pa = pa->next;pb = pb->next;}while(pa!= NULL){node *t = new node;t->data = pa->data + step;step = 0;pi->next = t;pi = t;pa = pa->next; }while(pb != NULL){node *t = new node;t->data = pb->data + step;step = 0;pi->next = t;pi = t;pb = pa->next;}if (step>0){node *t = new node;t->data = step;pi->next = t;pi = t;}pi->next = NULL;return pc;}int main()
{int a[6] = {1,4,5,9,7,8};node *head = new node;init(head,a,6);print(head);int b[6] = {1,4,5,9,7,8};node *h = new node;init(h,a,6);print(h);node * r = plus(head, h);print(r);return 0;
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