2. Add Two Numbers

1. 问题描述

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

Tags: Linked List Math
Similar Problems: (M) Multiply Strings (E) Add Binary

2. 解答思路

3. 代码

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
#include <iostream>
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (NULL == l1)
{
return l2;
}
if (NULL == l2)
{
return l1;
}
return addTwoNum(l1, l2);
}
public:
void PrintList(ListNode* l1)
{
if (NULL == l1)
{
return;
}
ListNode *pt = l1;
while (NULL != pt)
{
std::cout<< pt->val << " ";
pt = pt->next;
}

}
private:
ListNode* addTwoNum(ListNode* ptLar, ListNode* ptSma)
{
ListNode *pLar = ptLar;
ListNode *pSma = ptSma;
int t = 0;
while (NULL != pSma && NULL != pLar)
{
int sum = pLar->val + pSma->val + t;
pLar->val = sum%10;
pSma->val = pLar->val;
t = sum/10;
if ((NULL != pSma->next && NULL != pLar->next))
{
pSma = pSma->next;
pLar = pLar->next;
}
else
{
break;
}

}

if (0 == t)
{
if (NULL == pSma->next)
{
return ptLar;
}
else
{
return ptSma;
}
}

if (NULL == pSma->next && NULL == pLar->next)
{
ListNode *pNew = new ListNode(1);
pLar->next = pNew;
return ptLar;
}

if ( NULL != pLar->next)
{
addone(pLar);
return ptLar;
}
else
{
addone(pSma);
return ptSma;
}

}
void addone(ListNode *pLar, int t = 1)
{
pLar = pLar->next;
while (NULL != pLar)
{
int sum = pLar->val + t;
pLar->val = sum%10;
t = sum/10;
if (0 == t)
{
break;
}
else
{
if (NULL == pLar->next)
{
ListNode *pNew = new ListNode(1);
pLar->next = pNew;
break;
}
else
{
pLar = pLar->next;
}
}
}
}
};

4. 反思