-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathadd-two-numbers_20241005
43 lines (39 loc) · 1.53 KB
/
add-two-numbers_20241005
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
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<meta http-equiv="X-UA-Compatible" content="IE=edge" />
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<title>add-two-numbers</title>
</head>
<body></body>
<script>
/**
* @param {number} num
* @return {boolean}
* https://leetcode.com/problems/add-two-numbers/description/
*/
var addTwoNumbers = function (l1, l2) {
let dummyHead = new ListNode(0); // 虚拟头节点，帮助简化链表操作
let current = dummyHead; // 当前节点指针，初始指向虚拟头节点
let carry = 0; // 进位
// 当 l1 或 l2 中还有节点时，继续相加
while (l1 !== null || l2 !== null) {
const x = l1 !== null ? l1.val : 0; // 获取 l1 当前节点的值，若为空则为 0
const y = l2 !== null ? l2.val : 0; // 获取 l2 当前节点的值，若为空则为 0
const sum = carry + x + y; // 当前位的和
carry = Math.floor(sum / 10); // 计算进位
current.next = new ListNode(sum % 10); // 创建新节点存储当前位的结果
current = current.next; // 移动到下一个节点
// 移动到链表的下一个节点
if (l1 !== null) l1 = l1.next;
if (l2 !== null) l2 = l2.next;
}
// 如果最后还有进位，创建一个新的节点
if (carry > 0) {
current.next = new ListNode(carry);
}
return dummyHead.next; // 返回结果链表的头节点
};
</script>
</html>