算法分析与设计——LeetCode Problem.2 Add Two Numbers

题目链接

问题详情

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

解题思路：

题目要求给出两个表示数字的链表，数字的位数在链表中从低到高依次连接。然后返回两个链表表示的数相加的结果。显然，可以新建一个链表来保存计算结果。实现代码如下

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {    ListNode* h1 = l1;ListNode* h2 = l2;ListNode* follow = new ListNode(0);ListNode* sum = follow;ListNode* head = follow;sum->next = NULL;while (h1 != NULL && h2 != NULL) {sum->val = (h1->val + h2->val) + sum->val;if (sum->val >= 10) {sum->val %= 10;sum->next = new ListNode(1);}else {sum->next = new ListNode(0);}h1 = h1->next;h2 = h2->next;follow = sum;//follow->next = NULL;sum = sum->next;sum->next = NULL;}if (h1 == NULL) {while (h2 != NULL) {sum->val = (h2->val + sum->val);if (sum->val >= 10) {sum->val = sum->val % 10;sum->next = new ListNode(1);//sum = sum->next;}else {sum->next = new ListNode(0);//sum = sum->next;}h2 = h2->next;follow = sum;sum = sum->next;sum->next = NULL;}//return head;if (follow->next->val == 0) {follow->next = NULL;}else {follow = follow->next;follow->next = NULL;}return head;}if (h2 == NULL) {while (h1 != NULL) {sum->val = (h1->val + sum->val);if (sum->val >= 10) {sum->val = sum->val % 10;sum->next = new ListNode(1);//sum = sum->next;}else {sum->next = new ListNode(0);//sum = sum->next;}h1 = h1->next;follow = sum;sum = sum->next;sum->next = NULL;}//return head;if (follow->next->val == 0) {follow->next = NULL;}else {follow = follow->next;follow->next = NULL;}return head;}    }};