leetcode

commit 3f1ca9c882f606d0ffcb5d94eda594f640f45961
parent be0a5d399a6820370f75e37b6adeb11b030b98c9
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Mon,  1 May 2023 11:49:18 -0500

Completed remove nodes from linked list

Diffstat:
A
remove-nodes-from-linked-list/remove-nodes-from-linked-list.dart
 | 
44
++++++++++++++++++++++++++++++++++++++++++++

1 file changed, 44 insertions(+), 0 deletions(-)
diff --git a/remove-nodes-from-linked-list/remove-nodes-from-linked-list.dart b/remove-nodes-from-linked-list/remove-nodes-from-linked-list.dart
@@ -0,0 +1,44 @@
+//Return a linked list of all nodes that do not contain a higher value on their right
+//To solve this I created an array of all the linked list items. I then iterated through it
+//from the left searching to see if a higher value on the right existed. If it did not then
+//it would link up to the last node found and the new node would be the head.
+//The time complexity of this code is O(n) where n is the number of nodes.
+//Time: 497ms Beats: 100%
+//Memory: 220.7MB Beats: 100%
+//This question did not have enough submissions to show
+//metrics, but I am confident my solution is rather fast.
+
+class Solution {
+  ListNode? removeNodes(ListNode? head) {
+
+      List<int> reverse_list = [];
+      while(head != null){
+          reverse_list.add(head.val);
+          head = head.next;
+      }
+      int highest = 0;
+      ListNode current = new ListNode();
+      ListNode? last;
+
+      for(int i = reverse_list.length - 1; i >= 0 ;--i){
+        if(reverse_list[i] >= highest){
+            highest = reverse_list[i];
+            current = new ListNode();
+            current.val = highest;
+            if(last != null){
+              current.next = last;
+              head = current;
+            }
+            else{
+              head = current;
+            }
+            last = current;
+          }
+      }
+
+
+
+
+      return head;
+  }
+}