LeetCode--Add Two Numbers

发表于 | 分类于 |

题目描述：

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.

样例：

1
2
3
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

题意：

给你两个表示非负数字的链表。数字以相反的顺序存储，其节点包含单个数字。将这两个数字相加并将其作为另外一个链表返回。

题解：

这道题其实就是一个暴力，时间的长短，主要就是看怎么一个写法，蚊子第一次的解法是这样的：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
/**
 * 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 *ans = new ListNode(0);
        ListNode *head = ans;
        int res = 0;
        while(l1 != NULL && l2 != NULL){
            int num = l1->val + l2->val + res;
            res = num / 10;
            num %= 10;
            ans->next = new ListNode(num);
            ans = ans->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1 != NULL){
            int num = l1->val + res;
            res = num / 10;
            num %= 10;
            ans->next = new ListNode(num);
            ans = ans->next;
            l1 = l1->next;
        }
        while(l2 != NULL){
            int num = l2->val + res;
            res = num / 10;
            num %= 10;
            ans->next = new ListNode(num);
            ans = ans->next;
            l2 = l2->next;
        }
        if(res > 0){
            ans->next = new ListNode(res);
        }
        return head->next;
    }
};

Runtime：40ms Memory：19.2MB faster than 96.34%

康了一下网友的题解，大致都跟蚊子的一样，这里就不拖出来康了。