LeetCode in Kotlin

203. Remove Linked List Elements

Easy

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

Example 1:

Input: head = [1,2,6,3,4,5,6], val = 6

Output: [1,2,3,4,5]

Example 2:

Input: head = [], val = 1

Output: []

Example 3:

Input: head = [7,7,7,7], val = 7

Output: []

Constraints:

• The number of nodes in the list is in the range [0, 104].
• 1 <= Node.val <= 50
• 0 <= val <= 50

Solution

import com_github_leetcode.ListNode

/*
 * 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 removeElements(head: ListNode?, `val`: Int): ListNode? {
        val sentinel = ListNode(-1)
        sentinel.next = head
        var next = head
        var prev = sentinel
        while (next != null) {
            if (next.`val` == `val`) {
                prev.next = next.next
            } else {
                prev = next
            }
            next = next.next
        }
        return sentinel.next
    }
}

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