LeetCode in Kotlin

3217. Delete Nodes From Linked List Present in Array

Medium

You are given an array of integers nums and the head of a linked list. Return the head of the modified linked list after removing all nodes from the linked list that have a value that exists in nums.

Example 1:

Input: nums = [1,2,3], head = [1,2,3,4,5]

Output: [4,5]

Explanation:

Remove the nodes with values 1, 2, and 3.

Example 2:

Input: nums = [1], head = [1,2,1,2,1,2]

Output: [2,2,2]

Explanation:

Remove the nodes with value 1.

Example 3:

Input: nums = [5], head = [1,2,3,4]

Output: [1,2,3,4]

Explanation:

No node has value 5.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105
• All elements in nums are unique.
• The number of nodes in the given list is in the range [1, 105].
• 1 <= Node.val <= 105
• The input is generated such that there is at least one node in the linked list that has a value not present in nums.

Solution

import com_github_leetcode.ListNode
import kotlin.math.max

/*
 * 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 modifiedList(nums: IntArray, head: ListNode?): ListNode? {
        var maxv = 0
        for (v in nums) {
            maxv = max(maxv, v)
        }
        val rem = BooleanArray(maxv + 1)
        for (v in nums) {
            rem[v] = true
        }
        val h = ListNode(0)
        var t = h
        var p = head
        while (p != null) {
            if (p.`val` > maxv || !rem[p.`val`]) {
                t.next = p
                t = p
            }
            p = p.next
        }
        t.next = null
        return h.next
    }
}

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