链表求和

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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

思考

第一个数字作为表头，说明直接按顺序从左到有进行运算即可，需要考虑最后一个节点和的值
后面进行了改进，结果判定式中把 == 写成 = 号又折磨了我好久

代码

//  By Lentitude/** * 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        if (!l1) return l2;        if (!l2) return l1;        int numOfl1 = 0;        // 保存链表 l1 的结点数        int numOfl2 = 0;        // 保存链表 l2 的节点数        ListNode *temp1 = l1, *temp2 = l2;        while (temp1){            ++numOfl1;            temp1 = temp1->next;        }        while (temp2){            ++numOfl2;            temp2 = temp2->next;        }        ListNode *l3;        if (numOfl1 >= numOfl2){            l3 = result(l1, numOfl1, l2, numOfl2);        }else{            l3 = result(l2,numOfl2, l1, numOfl1);        }        return l3;    }    // 根据长度返回链表    /**     * parms l1: the longer list     * parm  l2: the short list     * return : the sum of l1 and l2     */    ListNode *result(ListNode *l1, int numOfl1, ListNode *l2, int numOfl2){        ListNode *l3 = l1;        int sign = 0;    // 表示进位        while (l1){           int temp = l1->val + l2->val + sign;            if (temp <=9){               l1->val = temp;               sign = 0;           }else{               l1->val = temp % 10;               if (l1->next == NULL){                   l1->next = new ListNode(temp/10);                   break;               }else{                   sign = temp / 10;               }           }           if (l2 ->next == NULL){               l2->next = new ListNode(0);           }           l1 = l1->next;           l2 = l2->next;           }        return l3;    }};

//  By Lentitude/** * 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        if (!l1) return l2;        if (!l2) return l1;        ListNode *temp3 = new ListNode(0);        ListNode *l3 = temp3;        int sign = 0;   // 进位        while (true)        {            int temp = l1->val + l2->val + sign;            if (temp <= 9) {                temp3->val = temp;                sign = 0;                if (l1->next == NULL && l2->next == NULL) {                    break;                }            }            else {                temp3->val = temp % 10;                if (l1->next == NULL && l2->next == NULL) {                    temp3->next = new ListNode(temp / 10);                    break;                }                else {                    sign = temp / 10;                }            }            if (temp3->next == NULL) {                temp3->next = new ListNode(0);                temp3 = temp3->next;            }            if (l1->next == NULL && l2->next != NULL) {                l1->next = new ListNode(0);            }            if (l1->next != NULL && l2->next == NULL) {                l2->next = new ListNode(0);            }            l1 = l1->next;            l2 = l2->next;        }        return l3;    }};

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