Solved! Leetcode 2487. Remove Nodes From Linked List

Description: Remove Nodes From Linked List

You are given the head of a linked list.

Remove every node which has a node with a greater value anywhere to the right side of it.

Return the head of the modified linked list.

Example 1

<strong>Input:</strong> head = [5,2,13,3,8]
<strong>Output:</strong> [13,8]
<strong>Explanation:</strong> The nodes that should be removed are 5, 2 and 3.
- Node 13 is to the right of node 5.
- Node 13 is to the right of node 2.
- Node 8 is to the right of node 3.

Input: head = [5,2,13,3,8]

Output: [13,8]

Explanation: The nodes that should be removed are 5, 2 and 3.

- Node 13 is to the right of node 5.

- Node 13 is to the right of node 2.

- Node 8 is to the right of node 3.

Example 2

<strong>Input:</strong> head = [1,1,1,1]
<strong>Output:</strong> [1,1,1,1]
<strong>Explanation:</strong> Every node has value 1, so no nodes are removed.

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

Output: [1,1,1,1]

Explanation: Every node has value 1, so no nodes are removed.

Constraints

The number of the nodes in the given list is in the range [1, 10⁵].
1 <= Node.val <= 10⁵

Solution

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    //use stack
    public ListNode removeNodes(ListNode head) {
        ArrayDeque<ListNode> stack = new ArrayDeque<>();
        ListNode p = head;
        while(p!=null){

            //if the stack is not empty or the current node val is greater than the previous node, pop an element from the stack
            while(!stack.isEmpty() && stack.peek().val < p.val){
                stack.pop();
            }
            stack.push(p);
            p = p.next;
        }

        p = stack.pop();

        //update the next pointer
        while(!stack.isEmpty()){
            ListNode n = stack.pop();
            n.next = p;
            p = n;
        }

        return p;
    }
}

/**

* Definition for singly-linked list.

* public class ListNode {

* int val;

* ListNode next;

* ListNode() {}

* ListNode(int val) { this.val = val; }

* ListNode(int val, ListNode next) { this.val = val; this.next = next; }

* }

class Solution {

//use stack

public ListNode removeNodes(ListNode head) {

ArrayDeque<ListNode> stack = new ArrayDeque<>();

ListNode p = head;

while(p!=null){

//if the stack is not empty or the current node val is greater than the previous node, pop an element from the stack

while(!stack.isEmpty() && stack.peek().val < p.val){

stack.pop();

}

stack.push(p);

p = p.next;

}

p = stack.pop();

//update the next pointer

while(!stack.isEmpty()){

ListNode n = stack.pop();

n.next = p;

p = n;

}

return p;

}

Time Complexity

O(n), where n is the number of nodes in a linkedlist

Space Complexity

O(n), where n is the number of nodes in a linkedlist

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