LeetCode-Algorithm2 C++的链表

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

这道题如果采用链表来做的话要考验一下细心程度，最后要判断一下是否有carryout，而不能仅以l1和l2是否为空来判断是否可以返回答案了。

还有C++中链表的节点支持构造器，比如：

struct ListNode{int val;ListNode *next;ListNode(int x) : val(x), next(NULL) {}};

就可以直接new ListNode(0)来构造一个val=0，next=NULL的新节点了
链表解法：

class Solution{int unit,tempsum;bool carryout=false;public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2){ListNode *pans=new ListNode(0);ListNode *ans = pans;if (l1->val + l2->val >= 10){carryout = true;pans->val = (l1->val + l2->val) % 10;}else{pans->val = l1->val + l2->val;carryout = false;}l1 = l1->next;l2 = l2->next;while (l1 != NULL||l2 != NULL){pans->next = new ListNode(0);pans = pans->next;if (l1 != NULL&&l2 != NULL){if (carryout == true){pans->val = 1;}tempsum = pans->val + l1->val + l2->val;if (tempsum >= 10){pans->val = tempsum % 10;carryout = true;}else{pans->val = tempsum;carryout = false;}l1 = l1->next;l2 = l2->next;}else if (l1 == NULL&&l2 != NULL){if (carryout == true){pans->val = 1;}tempsum = pans->val + l2->val;if (tempsum >= 10){pans->val = tempsum % 10;carryout = true;}else{pans->val = tempsum;carryout = false;}l2 = l2->next;}else if (l1 != NULL&&l2 == NULL){if (carryout == true){pans->val = 1;}tempsum = pans->val + l1->val;if (tempsum >= 10){pans->val = tempsum % 10;carryout = true;}else{pans->val = tempsum;carryout = false;}l1 = l1->next;}}if (carryout == true){pans->next = new ListNode(1);}return ans;}};

还有突然发现国外的OJ都是给定了接口，只让你实现一个函数，而不是全部代码都由你编写，做了那么多国内的OJ还是有点不习惯。用面向对象的语言写算法也确实没有面向过程那么习惯。