leetcode

commit e95e2099b07af4dd4d647795e3585fc9ca7636f5
parent 0a3c7b15a6db17c131535771145aa39ffe032294
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Sun, 14 May 2023 20:36:13 -0500

Completed swapping nodes in a linked list using dart

Diffstat:
A
swapping-nodes-in-a-linked-list/swapping-nodes-in-a-linked-list.dart
 | 
29
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1 file changed, 29 insertions(+), 0 deletions(-)
diff --git a/swapping-nodes-in-a-linked-list/swapping-nodes-in-a-linked-list.dart b/swapping-nodes-in-a-linked-list/swapping-nodes-in-a-linked-list.dart
@@ -0,0 +1,29 @@
+//Swap two nodes in a linked list without swapping the values.
+//Basically place the kth element in the length - kth spot
+//and vice versa not just swap the stored values.
+//I did this using an array list where I swapped the nodes
+//and then went through the list setting the next node to be
+//n + 1. This was the easiest solution I could think of.
+//The time complexity of this is O(n) where n is the lenght of the
+//list.
+//Time: 465ms Beats: 50%
+//Memory: 212.2MB Beats: 25%
+class Solution {
+  ListNode? swapNodes(ListNode? head, int k) {
+    ListNode? itr = head;
+    List<ListNode?> vals = [];
+    while(itr != null){
+      vals.add(itr);
+      itr = itr.next;
+    }
+    
+    ListNode? temp = vals[k - 1];
+    vals[k - 1] = vals[vals.length - k];
+    vals[vals.length - k] = temp;
+    for(int i = 0 ; i < vals.length - 1 ; ++i){
+      vals[i]?.next = vals[i + 1];
+    }
+    vals[vals.length - 1]?.next = null;
+    return vals[0];
+  }
+}