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👨🏿💻🙌.