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

#include <iostream>
using namespace std;

/**
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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

*/

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;
ListNode *l3 = new ListNode(-1);
ListNode *tnode = l3;
int over = 0;
while(l1 && l2)
{
int sum = l1->val + l2->val + over;
ListNode *node = new ListNode(sum % 10);
over = sum / 10;
tnode->next = node;
tnode = tnode->next;
l1 = l1->next;
l2 = l2->next;
}
if(l1 == NULL && l2 == NULL && over)//后一个节点，要考虑有没进位
{
ListNode *node = new ListNode(over);
tnode->next = node;
return l3->next;
}

ListNode *left = l1;
if(l2)
left = l2;
while(left)
{
int sum = left->val + over;
ListNode *node = new ListNode(sum % 10);
over = sum / 10;
tnode->next = node;
tnode = tnode->next;
left = left->next;

}
if(over)//同样，最后一个节点，要考虑有没进位
{
ListNode *node = new ListNode(over);
tnode->next = node;
}
return l3->next;

}

};
int main(void)
{
return 0;
}
posted @ 2014-04-29 14:07  mickole  阅读(318)  评论(0编辑  收藏  举报