2. Add Two Numbers

不是很难的一道题，就是list操作，但是细节总是写不对。

1 循环时是先while(l1 != NULL && l2 != NULL)然后在加剩下的list，但是发现这个时候也要考虑进位，所以把长list的后半段留下来并不能直接加在返回list的后面，所以还是用while(l1 != NULL || l2 != NULL)把两个list都遍历完，再处理剩下的情况，

2 但是使用sum来记录总和，而不是单记录carry再多存一个res，这样会更简洁也节省空间。

3 我最开始的时候在开头判断了一个 if (l1->val == 0） return l2; if (l2->val == 0) return l1; 因为想判断一个为0的情况，这样可以直接返回另一个list，但是忘记了list是逆序，第一个数为0，并不代表这个list表示的是0.

4 不要用ListNode* head = NULL, end = NULL; 这样会把end变成ListNode类型而不是ListNode*

最开始的版本：

/** * 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) {        ListNode* head = NULL;         ListNode* end = NULL;// don't use ListNode* head = NULL, end = NULL; will define end as ListNode instead of ListNode*        int carry = 0;        while (l1 != NULL && l2 != NULL) {            int res = l1->val + l2->val + carry;            ListNode* node;            if (res > 9) {                node = new ListNode(res % 10);                cout << node->val << endl;            }            else {                node= new ListNode(res);                cout << node->val <<endl;            }            if (head == NULL) {                head = node;                end = node;            }            else {                end->next = node;                end = end->next;            }            carry = res/10;            l1 = l1->next;            l2 = l2->next;        }        while (l1 != NULL) {            int res = l1->val + carry;            ListNode* node;            if (res > 9) {                node = new ListNode(res % 10);            }            else {                node = new ListNode(res);            }            if (head == NULL) {                head = node;                end = node;            }            else {                end->next = node;                end = end->next;            }            carry = res/10;            l1 = l1->next;        }        while (l2 != NULL) {            int res = l2->val + carry;            ListNode* node;            if (res > 9) {                node = new ListNode(res % 10);            }            else {                node = new ListNode(res);            }            if (head == NULL) {                head = node;                end = node;            }            else {                end->next = node;                end = end->next;            }            carry = res/10;            l2 = l2->next;        }        if (carry) {//don't forget to add this node when carry is not 0            ListNode* node = new ListNode(carry);           end->next = node;            end = end->next;        }        return head;    }};

修改后更简洁的版本：

/** * 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) {        ListNode* head = NULL;         ListNode* end = NULL;        int sum = 0;        while (l1 != NULL || l2 != NULL) {            if (l1 != NULL) {                sum += l1->val;                l1 = l1->next;            }            if (l2 != NULL) {                sum += l2->val;                l2 = l2->next;            }             ListNode* node = new ListNode(sum%10);            sum = sum / 10;            if (head == NULL) {                head = node;                end = node;            }            else {                end->next = node;                end = end->next;            }        }        while (sum) {            ListNode* node = new ListNode(sum%10);            end->next = node;            end = end->next;            sum = sum /10;        }        return head;    }};