Leetcode | Add Two Numbers

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

注意取指针时要判断是否为空。

一般尾插入的话，用一个dummy结点。这样就可以不用特殊处理头指针了。

 1 /**

 2  * Definition for singly-linked list.

 3  * struct ListNode {

 4  *     int val;

 5  *     ListNode *next;

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

 7  * };

 8  */

 9 class Solution {

10 public:

11     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

12         ListNode ret(0);

13         ListNode *p = &ret;

14         int carry = 0, v1, v2, s;

15         

16         while (l1 != NULL || l2 != NULL) {

17             v1 = l1 ? l1->val : 0;

18             v2 = l2 ? l2->val : 0;

19             s = v1 + v2 + carry;

20             carry = s / 10;

21             p->next = new ListNode(s % 10);

22             p = p->next;

23             if (l1) l1 = l1->next;

24             if (l2) l2 = l2->next;

25         }

26         if (carry > 0) {

27             p->next = new ListNode(carry);

28         }

29         return ret.next;

30     }

31 };

 代码现在是越写越简洁了，好事。逻辑清晰了许多。

 1 /**

 2  * Definition for singly-linked list.

 3  * struct ListNode {

 4  *     int val;

 5  *     ListNode *next;

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

 7  * };

 8  */

 9 class Solution {

10 public:

11     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

12         int carry = 0, v = 0;

13         ListNode h(0), *p = &h;

14         while (l1 || l2) {

15             v = carry;

16             if (l1) {

17                 v += l1->val;

18                 l1 = l1->next;

19             } 

20             if (l2) {

21                 v += l2->val;

22                 l2 = l2->next;

23             }

24             p->next = new ListNode(v % 10);

25             p = p->next;

26             carry = v / 10;

27         }

28         if (carry > 0) {

29             p->next = new ListNode(carry);

30         }

31         return h.next;

32     }

33 };