标签：des style blog http io ar color os sp
题目描述：
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
起初，这道题目被我想简单了。我用两个int值分别保存两个链表所表示的值，然后计算它们的和，最后把结果转换为链表。然而链表所能表示的值可以无限大，超出了int的存储范围。
后来自己思索了下，写出了以下代码：
solution1：
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { int first = 0; int second = 0; int carry = 0; int sum = 0; ListNode *list = new ListNode(-1); ListNode *p = list; while (l1 && l2) { first = l1->val; second = l2->val; sum = (first + second + carry) % 10; if(first + second + carry >= 10) carry = 1; else carry = 0; p->next = new ListNode(sum); p = p->next; l1 = l1->next; l2 = l2->next; } while (l1) { first = l1->val; if (carry) { sum = (first + carry) % 10; if(first + carry >= 10) carry = 1; else carry = 0; } else { sum = first; } p->next = new ListNode(sum); p = p->next; l1 = l1->next; } while (l2) { second = l2->val; if (carry) { sum = (second + carry) % 10; if(second + carry >= 10) carry = 1; else carry = 0; } else { sum = second; } p->next = new ListNode(sum); p = p->next; l2 = l2->next; } if (carry) { p->next = new ListNode(1); } return list->next; }
代码比较冗余，且可读性不高。后来在论坛上找到了别人的优化算法，真的很赞。
solution2：
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode *p1 = l1; ListNode *p2 = l2; ListNode *list = new ListNode(0); ListNode *p = list; int sum = 0; while (p1 != NULL || p2 != NULL) { if (p1 != NULL) { sum += p1->val; p1 = p1->next; } if (p2 != NULL) { sum += p2->val; p2 = p2->next; } p->next = new ListNode(sum % 10); p = p->next; sum /= 10; } if(sum) p->next = new ListNode(1); return list->next; }
原文链接：https://oj.leetcode.com/discuss/2308/is-this-algorithm-optimal-or-what
ps:
代码AC只是第一步，要精益求精！
标签：des style blog http io ar color os sp
原文地址：http://www.cnblogs.com/gattaca/p/4157565.html
