leetcode

commit 35407ecf90f4b048329122f7e13251ab0b7cbcc4
parent 726c43dd49b4408de500d11ea32b8202f891ceaa
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Tue,  2 May 2023 11:12:44 -0500

Completed rotate-list using dart

Diffstat:
A
rotate-list/rotate-list.dart
 | 
57
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

1 file changed, 57 insertions(+), 0 deletions(-)
diff --git a/rotate-list/rotate-list.dart b/rotate-list/rotate-list.dart
@@ -0,0 +1,57 @@
+//Rotate a list k times to the right and return the rotated linked list.
+//To solve this I found the remainder of the k / length of linked list 
+//then I iterated through the output of that each time moving all of the values one
+//to the right in the list.
+//Time: 280ms Beats: 100%
+//Memory: 144.3MB Beats: 100%
+//This problem did not have enough submissions for dart to give accuarte metrics.
+//To optimize my solution I would iterate through the list one time first finding all of the values
+//that overflowed to the start then adding on the rest of the values. That solution would be O(n) time as opposed
+//to my solution which is O(n*k)
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *   int val;
+ *   ListNode? next;
+ *   ListNode([this.val = 0, this.next]);
+ * }
+ */
+class Solution {
+  ListNode? rotateRight(ListNode? head, int k) {
+      if(head == null){
+          return head;
+      }
+      List<int> vals = [];
+      while(head != null){
+          vals.add(head.val);
+          head = head.next;
+      }
+      k = k % vals.length;
+
+
+
+      List<int> original =  List.from(vals);
+      for(int x = 0 ; x < k ; ++x){
+        for(int i = 1 ; i < vals.length ; ++i){
+            vals[i] = original[i - 1];
+        }
+        vals[0] = original[original.length - 1];
+        original =  List.from(vals);
+      }
+      ListNode? start;
+      ListNode last = new ListNode();
+      for(int i = 0 ; i < vals.length ; ++i){
+          ListNode current = new ListNode();
+          current.val = vals[i];
+          if(start == null){
+              start = current;
+              last = current;
+          }
+          else{
+              last.next = current;
+              last = current;
+          }
+      }
+      return start;
+    }
+}