Add Two Numbers

题目来源：

https://leetcode.com/problems/add-two-numbers/description/

题目描述：

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

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

心路历程：

这道题在leetcode上的难度标示为Medium，但感觉不是太难，主要考查结构体和链表的运用。其中有些特殊情况，是在提交之后报错才知道的。比如，要考虑最高位的进位。还有一个关键点，就是两个链表长度不一致时，要把比较短的那个链表高位用0来代替。

解决方案：

1.自己写的C++版解法，遇到的一个坑是，结构体等于NULL的时候，不能再指向下一个节点。（一开始天真地以为NULL的下一个节点还是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) {ListNode* head = new ListNode(0);ListNode* result = head;int flag = 0;int add1, add2, sum, rem;//加数1，加数2，和，余数int carry = 0;//进位while (l1 != NULL || l2 != NULL) {//只要任意一个链表不为空则继续if (l1 == NULL) add1 = 0; else add1 = l1->val;//空的链表赋值为0if (l2 == NULL) add2 = 0; else add2 = l2->val;sum = add1 + add2 + carry;carry = sum / 10;rem = sum % 10;if (flag == 0) { //链表头直接赋值head->val = rem;flag++;}else { //新建节点head->next = new ListNode(rem);head = head->next;}if (l1 != NULL)l1 = l1->next;if (l2 != NULL) l2 = l2->next;}if (carry == 0){}//处理最高位else {head->next = new ListNode(carry);head = head->next;}return result;}};int main() {ListNode* l1 = new ListNode(0);ListNode* l2 = new ListNode(0);ListNode* arg1 = l1;ListNode* arg2 = l2;l1->val = 2;l1->next = new ListNode(4);l1 = l1->next;l1->next = new ListNode(3);l2->val = 5;l2->next = new ListNode(6);l2 = l2->next;l2->next = new ListNode(4);Solution solution;ListNode* result = solution.addTwoNumbers(arg1, arg2);while (result != NULL) {cout << result->val;result = result->next;}cout << endl;system("pause");}

2.第二种解决方案来自leetcode，用Java写的，思路与第一种基本一致。主要的优化是采用了虚拟链表头，使得循环时不需要进行一次判断，最后返回链表头的下一个节点即可。这种思路值得借鉴。代码如下：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {    ListNode dummyHead = new ListNode(0);    ListNode p = l1, q = l2, curr = dummyHead;    int carry = 0;    while (p != null || q != null) {        int x = (p != null) ? p.val : 0;        int y = (q != null) ? q.val : 0;        int sum = carry + x + y;        carry = sum / 10;        curr.next = new ListNode(sum % 10);        curr = curr.next;        if (p != null) p = p.next;        if (q != null) q = q.next;    }    if (carry > 0) {        curr.next = new ListNode(carry);    }    return dummyHead.next;