LeetCode 445. Add Two Numbers II

Solution1：我的答案
利用了栈，这样就不用翻转链表了。。。

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { //借助两个栈来运算if (l1 == NULL) return l2;else if (l2 == NULL)return l1;int carry = 0;stack<struct ListNode*> List1, List2, res_stack;struct ListNode* temp1 = l1, * temp2 = l2, * res = new ListNode(-1);struct ListNode* temp_res = res;while (temp1 || temp2) {if (temp1) {List1.push(temp1);temp1 = temp1->next;}if (temp2) {List2.push(temp2);temp2 = temp2->next;}}int n1 = 0, n2 = 0;while (!List1.empty() || !List2.empty()) {if (!List1.empty()) {n1 = List1.top()->val;List1.pop();}else if(List1.empty())n1 = 0;if (!List2.empty()) {n2 = List2.top()->val;List2.pop();}else if(List2.empty())n2 = 0;int sum = n1 + n2 + carry;carry = sum/10;res_stack.push(new ListNode(sum%10));}if (carry) res_stack.push(new ListNode(carry));while (!res_stack.empty()) {temp_res->next = res_stack.top();res_stack.pop();//原来stack的pop()操作并没有释放内存，虚惊一场~~~temp_res = temp_res->next;}return res->next;}
};

Solution2：
前半段思路相同，后半段思路要比我的好很多。。。实现的代码也大大优化了。。
参考网址：http://www.cnblogs.com/grandyang/p/6216480.html
思路：
我们首先遍历两个链表，将所有数字分别压入两个栈s1和s2中，我们建立一个值为0的res节点，然后开始循环，如果栈不为空，则将栈顶数字加入sum中，然后将res节点值赋为sum%10，然后新建一个进位节点head，赋值为sum/10，如果没有进位，那么就是0，然后我们head后面连上res，将res指向head，这样循环退出后，我们只要看res的值是否为0，为0返回res->next，不为0则返回res即可，参见代码如下：

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {stack<int> s1, s2;while (l1) {s1.push(l1->val);l1 = l1->next;}while (l2) {s2.push(l2->val);l2 = l2->next;}int sum = 0;ListNode *res = new ListNode(0);while (!s1.empty() || !s2.empty()) {if (!s1.empty()) {sum += s1.top(); s1.pop();}if (!s2.empty()) {sum += s2.top(); s2.pop();}res->val = sum % 10;//为当前结点赋值ListNode *head = new ListNode(sum / 10);//以进位值初始化新建的前一个结点，这便于while执行完head->next = res;//后进行判断。新建的结点串在当前结点的前面。res = head; //结点更新，把新建的结点更新为下次循环时的当前结点sum /= 10;  //sum保留了此次循环的进位信息，喵喵喵！}return res->val == 0 ? res->next : res;}
};

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