[LeetCode] 2. Add Two Numbers

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mb5fed4c003aebe 2019-10-13 06:41:00

文章标签 leetcode linked list java javascript git 文章分类 后端开发

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.

Example:

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

两数相加。

这题不难，思路是直接做加法，因为linked list给你的时候头结点是数字的最低位，不需要做reverse之类的操作，记得最后判断是不是还有额外的一个进位。

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {ListNode} l1
 3  * @param {ListNode} l2
 4  * @return {ListNode}
 5  */
 6 var addTwoNumbers = function (l1, l2) {
 7     let dummy = new ListNode(0);
 8     let cur = dummy;
 9     let carry = 0;
10     let p1 = l1;
11     let p2 = l2;
12     while (p1 !== null || p2 !== null) {
13         if (p1 !== null) {
14             carry += p1.val;
15             p1 = p1.next;
16         }
17         if (p2 !== null) {
18             carry += p2.val;
19             p2 = p2.next;
20         }
21         cur.next = new ListNode(carry % 10);
22         carry = parseInt(carry / 10);
23         cur = cur.next;
24     }
25     if (carry === 1) {
26         cur.next = new ListNode(1);
27     }
28     return dummy.next;
29 };

Java实现

 1 class Solution {
 2     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
 3         ListNode dummy = new ListNode(0);
 4         ListNode cur = dummy;
 5         int sum = 0;
 6         ListNode p1 = l1;
 7         ListNode p2 = l2;
 8         while (p1 != null || p2 != null) {
 9             if (p1 != null) {
10                 sum += p1.val;
11                 p1 = p1.next;
12             }
13             if (p2 != null) {
14                 sum += p2.val;
15                 p2 = p2.next;
16             }
17             cur.next = new ListNode(sum % 10);
18             sum /= 10;
19             cur = cur.next;
20         }
21         if (sum == 1) cur.next = new ListNode(1);
22         return dummy.next;
23     }
24 }