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odd-even-linked-list.dart (1274B)

      1 //Given a linked list return a new linked list that
      2 //is ordered with the odd indexes at the start and even indexes at
      3 //the end. To do this I created two seperate lists one for even one for odd.
      4 //I then passed them to a function to return a linked list based on the input
      5 //array.
      6 //The time complextiy is O(n) where n is the length of the linked list.
      7 //Time: 241ms Beats: 100%
      8 //Memory: 142.9MB Beats: 75%
      9 class Solution {
     10   ListNode? oddEvenList(ListNode? head) {
     11     if(head == null){
     12       return head;
     13     }
     14       List<int> even = [];
     15       List<int> odd = [];
     16       while(head != null){
     17           even.add(head.val);
     18           head = head.next;
     19           if(head?.val != null){
     20             odd.add((head?.val ?? 0));
     21           }
     22           head = head?.next;
     23       }
     24       return listify(even, odd);
     25   }
     26   ListNode listify(List<int> first , List<int> second){
     27     ListNode first_ln = new ListNode(first[0]);
     28     ListNode last = first_ln;
     29     for(int i = 1 ; i < first.length ; ++i){
     30       ListNode current = new ListNode(first[i]);
     31       last.next = current;
     32       last = current;
     33     }
     34     for(int i = 0 ; i < second.length ; ++i){
     35       ListNode current = new ListNode(second[i]);
     36       last.next = current;
     37       last = current;
     38     }
     39 
     40     return first_ln;
     41   }
     42 }