Leetcode日记(1)

1.题目

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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

img

1
2
3
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

1
2
Input: l1 = [0], l2 = [0]
Output: [0]

Example 3:

1
2
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:

  • The number of nodes in each linked list is in the range [1, 100].
  • 0 <= Node.val <= 9
  • It is guaranteed that the list represents a number that does not have leading zeros.

总的来说就是给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照逆序的方式存储的，并且每个节点只能存储一位数字。将两个数相加，并以相同形式返回一个表示和的链表。

2.解题思路

  1. 两个链表分别取数然后相加相加，若某一链表已经为空，则该链表对应的加数置为0。
  2. 根据和数得到相加后的本位和进位
  3. 直到历遍两个链表
  4. 如果进位还是非0，再在链表最后添加一个新的节点

3.代码实现

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
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int carry = 0;
        ListNode* l3 = new ListNode();
        ListNode* p = l3;
        
        while (l1 != nullptr || l2 != nullptr) {
            int a1 = (l1 != nullptr) ? l1->val : 0;
            int a2 = (l2 != nullptr) ? l2->val : 0;
            
            int sum = a1 + a2 + carry;
            carry = sum / 10;
            int digit = sum % 10;
            
            p->next = new ListNode(digit);
            p = p->next;
            
            if (l1 != nullptr) l1 = l1->next;
            if (l2 != nullptr) l2 = l2->next;
        }
        
        if (carry > 0) {
            p->next = new ListNode(carry);
        }
        
        return l3->next;
    }
};

4.输出结果

测试用例全部accept

image

忽略内存占用