[LeetCode] 2. Add Two Numbers（两数相加【链表版】）

• Difficulty: Medium

• Related Topics: Linked List, Math

• Link: https://leetcode.com/problems/add-two-numbers/

Description

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.
给定两个非空链表表示两个整数。整数的每一位倒序存储，链表内的每一个节点仅包含一个数字，将这两个数相加，以单链表的形式返回答案。

You may assume the two numbers do not contain any leading zero, except the number 0 itself.
你可以假定除了 0 本身外，两数都不含前导 0。

Examples

Example 1

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

Example 2

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

Example 3

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.

Solution

做这题的时候需要注意两点：

• 两数长度不一样时的处理；

• 相加到最后的进位处理。

得益于 Kotlin 对可空元素的语法糖，我不需要单独考虑第一种情况，代码如下：

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun addTwoNumbers(l1: ListNode?, l2: ListNode?): ListNode? {
        var p = l1
        var q = l2
        val dummy = ListNode(-1)
        var r: ListNode? = dummy
        var carry = 0

        while (p != null || q != null) {
            val digit = (p?.`val`?:0) + (q?.`val`?:0) + carry
            carry = digit / 10
            r?.next = ListNode(digit % 10)
            r = r?.next
            p = p?.next
            q = q?.next
        }
        
        if (carry != 0) {
            r?.next = ListNode(carry)
        }

        return dummy.next
    }
}