Leetcode Problem 2: Add Two numbers

Description

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

这题的意思是用链表表示两个逆序数，将他们相加后输出链表表示逆序的和。

Solution

这题主要考察链表的使用，其实很简单，从低位到高位遍历（链表中表现为从头指针开始向后遍历）。遍历结束的条件是两个链表节点都已经指向NULL。两个链表对应节点位置的数字依次相加，这个时候可能会产生进位。如果产生进位，则存储这个进位并加入到下一次计算中，直到两个链表均遍历完，输出结果。

Code

/**
* Leetcode problem list 2: add two number.
*
*
* hellfire(asyncloadng#163.com)
* Feb 15th, 2016
*/
#include<iostream>
using namespace std;

struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

/**
* Definition for singly-linked list.
* 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)
{
return l2;
}
if (l2 == NULL)
{
return l1;
}
ListNode *p, *l3;
p = new ListNode(-1);
int carry = 0;
l3 = p;

while (l1 != NULL || l2 != NULL)
{
if (l1 != NULL)
{
carry += l1 -> val;
l1 = l1 -> next;
}
if (l2 != NULL)
{
carry += l2 -> val;
l2 = l2 -> next;
}
l3 -> next = new ListNode(carry % 10);
l3 = l3 -> next;
carry = carry / 10;
}

if (carry == 1)
{
l3 -> next = new ListNode(1);
}
return p -> next;
}
};

int main(int argc, char **argv)
{
///////  Test case.
///////  Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
///////  Output: 7 -> 0 -> 8
Solution s;
ListNode *l1, *l2, *l3;
l1 = new ListNode(5);
l2 = new ListNode(5);
l3 = s.addTwoNumbers(l1, l2);
cout << l3 -> val << endl;
}

源代码对应的Github地址：https://github.com/oj-problem/leetcode/blob/master/solution2.cpp

备注

测试用例：

[0] [0] -> [0]

[1] [0] -> [1]

[1] [1] -> [2]

[5] [5] -> [1][0]