Fix analyze errors

nullskill · nullskill · commit a689d1f1faf8 · 2023-09-23T16:12:09.000+03:00
diff --git a/lib/src/easy/21.merge_two_sorted_lists/main.dart b/lib/src/easy/21.merge_two_sorted_lists/main.dart
@@ -7,14 +7,12 @@
 
 import '../../structure/list_node.dart';
 
-/**
- * Definition for singly-linked list.
- * class ListNode {
- *   int val;
- *   ListNode? next;
- *   ListNode([this.val = 0, this.next]);
- * }
- */
+/// Definition for singly-linked list.
+/// class ListNode {
+///   int val;
+///   ListNode? next;
+///   ListNode([this.val = 0, this.next]);
+/// }
 class Solution {
   ListNode? mergeTwoLists(ListNode? list1, ListNode? list2) {
     ListNode dummy = ListNode(0);
diff --git a/lib/src/easy/83.remove_duplicates_from_sorted_list/main.dart b/lib/src/easy/83.remove_duplicates_from_sorted_list/main.dart
@@ -9,14 +9,12 @@
 
 import 'package:leetcode/src/structure/list_node.dart';
 
-/**
- * Definition for singly-linked list.
- * class ListNode {
- *   int val;
- *   ListNode? next;
- *   ListNode([this.val = 0, this.next]);
- * }
- */
+/// Definition for singly-linked list.
+/// class ListNode {
+///   int val;
+///   ListNode? next;
+///   ListNode([this.val = 0, this.next]);
+/// }
 class Solution {
   ListNode? deleteDuplicates(ListNode? head) {
     if (head == null) return null;
diff --git a/lib/src/easy/94.binary_tree_inorder_traversal/main.dart b/lib/src/easy/94.binary_tree_inorder_traversal/main.dart
@@ -3,15 +3,13 @@
 
 import 'package:leetcode/src/structure/tree_node.dart';
 
-/**
- * Definition for a binary tree node.
- * class TreeNode {
- *   int val;
- *   TreeNode? left;
- *   TreeNode? right;
- *   TreeNode([this.val = 0, this.left, this.right]);
- * }
- */
+/// Definition for a binary tree node.
+/// class TreeNode {
+///   int val;
+///   TreeNode? left;
+///   TreeNode? right;
+///   TreeNode([this.val = 0, this.left, this.right]);
+/// }
 class Solution {
   List<int> inorderTraversal(TreeNode? root) {
     List<int> res = [];
diff --git a/lib/src/medium/49.group_anagrams/main.dart b/lib/src/medium/49.group_anagrams/main.dart
@@ -38,19 +38,19 @@ void main(List<String> args) {
 /// The time complexity of this variant is `O(n * k * log(k))`, where `n` is the length of
 /// the input list `strs` and `k` is the maximum length of a string in `strs`.
 
-List<List<String>> _groupAnagrams(List<String> strs) {
-  Map<String, List<String>> anagramGroups = {};
-
-  for (String str in strs) {
-    String sortedStr = String.fromCharCodes(str.runes.toList()..sort());
-    if (!anagramGroups.containsKey(sortedStr)) {
-      anagramGroups[sortedStr] = [];
-    }
-    anagramGroups[sortedStr]?.add(str);
-  }
-
-  return anagramGroups.values.toList();
-}
+// List<List<String>> _groupAnagrams(List<String> strs) {
+//   Map<String, List<String>> anagramGroups = {};
+
+//   for (String str in strs) {
+//     String sortedStr = String.fromCharCodes(str.runes.toList()..sort());
+//     if (!anagramGroups.containsKey(sortedStr)) {
+//       anagramGroups[sortedStr] = [];
+//     }
+//     anagramGroups[sortedStr]?.add(str);
+//   }
+
+//   return anagramGroups.values.toList();
+// }
 
 /// NB: The sorting variant is slower than counting the frequency of each character.
 /// However, the second algorithm may be faster than the first algorithm for inputs 
