2487. Remove Nodes From Linked List

In this post i'm gone explain liked list an famous leetcode problem that is "Remove Nodes from linked list".

Problem Statement:
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:

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.

Explanation:
In this problem statement state that remove the nodes which have the right side (any place) element greater than. let's understand with given example.

• Node 13 is the right side of the 5,2 nodes thats why 2,5 should be remove.
• Node 8 is the right side of 3 node thats why 3 should be remove.

final result would be [13,8]

Solution of the problem:

`/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */

const reverList = function(head){
    let prev = null;
    let curr = head;
    let next = null;

    while(curr!=null){
        next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }

    return prev;
}

var removeNodes = function(head) {
    // reverse list
    let reversList = reverList(head);

    let maxNode =  reversList;
    let prevNode = reversList;
    let currNode = reversList.next;
    // removed list
    while(currNode != null){
        if(maxNode.val > currNode.val){
            currNode = currNode.next;
        }else{
            maxNode = currNode;
            prevNode.next = currNode;
            prevNode = prevNode.next;
            currNode = currNode.next;
        }
    }

    prevNode.next = null;
    // reverse list
    return reverList(reversList);
};`

If you have any query or suggestions leave your expression👨🏿‍💻🙌.