leetcode

commit 3da4ee0462ca7e1d27aaf27c49a4414785168a25
parent 042885677c6be16c19b999dbe38da52ba182e253
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Wed,  3 May 2023 15:00:09 -0500

Completed reverse linked list ii problem using dart

Diffstat:
A
reverse-linked-list-ii/reverse-linked-list-ii.dart
 | 
52
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1 file changed, 52 insertions(+), 0 deletions(-)
diff --git a/reverse-linked-list-ii/reverse-linked-list-ii.dart b/reverse-linked-list-ii/reverse-linked-list-ii.dart
@@ -0,0 +1,52 @@
+//Reverse the nodes between the left and right points of a linked list.
+//To do this I put all of the nodes into an array and then using that array I created 
+//a subarray of all the values between the left and the right nodes. Then I iterated through
+//the list backwards to place them back into the original array. From there I turned the array
+//into a linked list and returned it. Despite the number of steps this algorithm is still O(n) time 
+//complexity where n is the number of nodes in the list.
+//Time: 260ms Beats: 80%
+//Memory: 142.7MB Beats: 80%
+
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *   int val;
+ *   ListNode? next;
+ *   ListNode([this.val = 0, this.next]);
+ * }
+ */
+class Solution {
+  ListNode? reverseBetween(ListNode? head, int left, int right) {
+      List<int> vals = [];
+      while(head != null){
+        vals.add(head.val);
+        head = head.next;
+      }
+      List<int> between_list = [];
+      int diff = right - left;
+      for(int i = left - 1 ; i <= left - 1 + diff ; ++i){
+        between_list.add(vals[i]);
+      }
+      int counter = 0;
+      for(int i = right - 1 ; i >= left - 1 ; --i){
+          vals[i] = between_list[counter];
+          counter += 1;
+      }
+      ListNode? last;
+      ListNode? first;
+      for(int i = 0 ; i < vals.length ; ++i){
+        if(first == null){
+            first = new ListNode();
+            first.val = vals[i];
+            last = first;
+        }
+        else{
+          ListNode current = new ListNode();
+          current.val = vals[i];
+          last?.next = current;
+          last = current;
+        }
+      }
+      return first;
+  }
+}