
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
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
if(l1==NULL && l2==NULL)
return NULL;
if (l1==NULL)
return l2;
if (l2==NULL)
return l1;
ListNode *head = NULL , *pre = NULL;
int c=0;
while( l1!=NULL || l2!=NULL || c!=0 )
{
if(l1!=NULL)
{
c += l1->val;
l1 = l1->next;
}
if(l2!=NULL)
{
c += l2->val;
l2 = l2->next;
}
ListNode *p = new ListNode(c%10);
if(head == NULL)
head = p;
else
pre->next = p;
pre = p;
c /= 10;
}
return head;
}
};#include<iostream>
using namespace std;
#define N 3
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 && l2==NULL)
return NULL;
if (l1==NULL)
return l2;
if (l2==NULL)
return l1;
ListNode *head = NULL , *pre = NULL;
int c=0;
while( l1!=NULL || l2!=NULL || c!=0 )
{
if(l1!=NULL)
{
c += l1->val;
l1 = l1->next;
}
if(l2!=NULL)
{
c += l2->val;
l2 = l2->next;
}
ListNode *p = new ListNode(c%10);
if(head == NULL)
head = p;
else
pre->next = p;
pre = p;
c /= 10;
}
return head;
}
};
ListNode *creatlist()
{
ListNode *head=NULL;
for(int i=0; i<N; i++)
{
int a;
cin>>a;
ListNode *p;
p = (ListNode*)malloc(sizeof(ListNode));
p->val = a;
p->next = head;
head = p;
}
return head;
}
int main()
{
ListNode *list1 = creatlist();
ListNode *list2 = creatlist();
Solution lin;
ListNode *outlist = lin.addTwoNumbers( list1,list2 );
for(int i=0; i<N; i++)
{
cout<<outlist->val;
outlist = outlist->next;
}
}原文地址:http://blog.csdn.net/keyyuanxin/article/details/43817511