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 };