LeetCode 2. Add Two Numbers[Medium]

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.

对于给定两个非空链表代表两个整数。链表中位数为逆序排列，每个节点值为0-9数字。将两个整数求和并作为链表返回。

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

错误解法：先整数求和，再存入链表。

问题：整数大小越界。

/** * 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) {        int a = 0;        int b = 0;        int sum = 0;        int count = 1;        ListNode *head, *p;        do        {            a += l1->val * count;            count = count * 10;            l1 = l1->next;        }while(l1);        count = 1;        do        {            b += l2->val * count;            count = count * 10;            l2 = l2->next;        }while(l2);        sum = a + b;        head = new ListNode(0);        //head->next = NULL;        p = head;        do        {            ListNode *r = new ListNode(0);            r->val = sum % 10;            r->next = NULL;            p->next = r;            p = p->next;            sum = sum / 10;        }while(sum);        return head->next;    }};

正确个人解法：

不足：代码不简洁

/** * 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) {        int carry = 0;        ListNode* head = new ListNode(0);        ListNode* p = head;        while (l1 != NULL || l2 != NULL)        {            int x = l1 != NULL ? l1->val : 0;            int y = l2 != NULL ? l2->val : 0;            int sum = carry + x + y;            carry = x + y + carry > 9 ? 1 : 0;            ListNode* r = new ListNode(0);            r->val = sum % 10;            r->next = NULL;            p->next = r;            p = p->next;            l1 = l1 != NULL ? l1->next : l1;            l2 = l2 != NULL ? l2->next : l2;        }        if (carry)        {            ListNode* r = new ListNode(0);            r->val = 1;            r->next = NULL;            p->next = r;            p = p->next;               }        return head->next;    }};

最简洁解法：

1. 定义简洁
2. 堆栈用法简洁；
3. 链表赋值简洁；

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {    ListNode preHead(0), *p = &preHead;    int extra = 0;    while (l1 || l2 || extra) {        int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;        extra = sum / 10;        p->next = new ListNode(sum % 10);        p = p->next;        l1 = l1 ? l1->next : l1;        l2 = l2 ? l2->next : l2;    }    return preHead.next;}