lintcode add-two-numbers 链表求和

问题描述

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。
样例
给出两个链表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null

笔记

在代码2中，要注意的时判断l1和l2是否到了尽头。

            if (l1 != NULL)                l1 = l1->next;            if (l2 != NULL)                l2 = l2->next;

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    /**     * @param l1: the first list     * @param l2: the second list     * @return: the sum list of l1 and l2      */    ListNode *addLists(ListNode *l1, ListNode *l2) {        // write your code here        int carry = 0;        ListNode *pre = new ListNode(0);        ListNode *tmp = pre;        while (l1 != NULL && l2 != NULL)        {            int num = l1->val + l2->val + carry;            l1 = l1->next;            l2 = l2->next;            tmp -> next = new ListNode(num % 10);            tmp = tmp -> next;            carry = num / 10;        }        while (l1 != NULL)        {            int num = l1->val + carry;            l1 = l1->next;            tmp -> next = new ListNode(num % 10);            tmp = tmp -> next;            carry = num / 10;        }        while (l2 != NULL)        {            int num = l2->val + carry;            l2 = l2->next;            tmp -> next = new ListNode(num % 10);            tmp = tmp -> next;            carry = num / 10;        }        if (carry == 1)        {            tmp->next = new ListNode(1);            tmp = tmp->next;        }        tmp->next = NULL;        return pre->next;    }};

第二次写。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    /**     * @param l1: the first list     * @param l2: the second list     * @return: the sum list of l1 and l2      */    ListNode *addLists(ListNode *l1, ListNode *l2) {        // write your code here        ListNode *beforeHead = new ListNode(0);        ListNode *pre = beforeHead;        int carry = 0;        while (l1 !=  NULL || l2 != NULL)        {            int v1 = (l1 == NULL) ? 0 : l1->val;            int v2 = (l2 == NULL) ? 0 : l2->val;            int tmp = v1 + v2 + carry;            ListNode *now = new ListNode(tmp%10);            carry = tmp / 10;            pre->next = now;            pre = now;            if (l1 != NULL)                l1 = l1->next;            if (l2 != NULL)                l2 = l2->next;        }        if (carry)        {            ListNode *now = new ListNode(1);            pre->next = now;            pre = now;        }        pre->next = NULL;        return beforeHead->next;    }};