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‎OddEvenLinkedList/odd_even_linked_list.dart

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/*
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-* 328. Odd Even Linked List *-
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Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.
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The first node is considered odd, and the second node is even, and so on.
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Note that the relative order inside both the even and odd groups should remain as it was in the input.
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You must solve the problem in O(1) extra space complexity and O(n) time complexity.
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Example 1:
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Input: head = [1,2,3,4,5]
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Output: [1,3,5,2,4]
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Example 2:
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Input: head = [2,1,3,5,6,4,7]
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Output: [2,3,6,7,1,5,4]
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Constraints:
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The number of nodes in the linked list is in the range [0, 104].
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-106 <= Node.val <= 106
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*/
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// Definition for singly-linked list.
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class ListNode {
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int val;
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ListNode? next;
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ListNode([this.val = 0, this.next]);
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}
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class A {
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ListNode? oddEvenList(ListNode? head) {
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if (head != null) {
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ListNode? odd = head, even = head.next, evenHead = even;
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while (even != null && even.next != null) {
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odd?.next = odd.next?.next;
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even.next = even.next?.next;
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odd = odd?.next;
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even = even.next;
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}
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odd?.next = evenHead;
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}
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return head;
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}
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}
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class B {
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ListNode? oddEvenList(ListNode? head) {
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if (head == null || head.next == null) {
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return head;
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}
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ListNode? evenList = ListNode(-1);
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ListNode? oddList = ListNode(-1);
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ListNode? currentEven = evenList;
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ListNode? currentOdd = oddList;
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ListNode? current = head;
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int i = 1;
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while (current != null) {
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if (i % 2 != 0) {
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currentOdd?.next = current;
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currentOdd = current;
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current = current.next;
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} else {
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currentEven?.next = current;
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currentEven = current;
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current = current.next;
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}
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i++;
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}
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currentOdd?.next = evenList.next;
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currentEven?.next = null;
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return oddList.next;
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}
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}
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class C {
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ListNode? oddEvenList(ListNode? head) {
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ListNode? lastInEven = null;
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ListNode? temp = head;
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while (head != null && head.next != null) {
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ListNode? next = head.next;
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if (lastInEven == null) lastInEven = next;
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ListNode? temp = lastInEven?.next;
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if (temp == null) break;
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head.next = temp;
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lastInEven?.next = temp.next;
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temp.next = next;
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lastInEven = lastInEven?.next;
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head = head.next;
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if (lastInEven == null) break;
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}
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return temp;
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}
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}

‎OddEvenLinkedList/odd_even_linked_list.go

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package main
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// Definition for singly-linked list.
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type ListNode struct {
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Val int
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Next *ListNode
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}
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func oddEvenList(head *ListNode) *ListNode {
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if head == nil || head.Next == nil {
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return head
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}
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var evenList *ListNode = &ListNode{Val: -1}
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var oddList *ListNode = &ListNode{Val: -1}
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var currentEven *ListNode = evenList
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var currentOdd *ListNode = oddList
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var current *ListNode = head
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var i int = 1
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for current != nil {
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if i%2 != 0 {
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currentOdd.Next = current
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currentOdd = current
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current = current.Next
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} else {
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currentEven.Next = current
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currentEven = current
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current = current.Next
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}
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i++
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}
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currentOdd.Next = evenList.Next
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currentEven.Next = nil
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return oddList.Next
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}
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/*
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ListNode? oddEvenList(ListNode? head) {
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if (head == null || head.next == null) {
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return head;
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}
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ListNode? evenList = ListNode(-1);
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ListNode? oddList = ListNode(-1);
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ListNode? currentEven = evenList;
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ListNode? currentOdd = oddList;
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ListNode? current = head;
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int i = 1;
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while (current != null) {
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if (i % 2 != 0) {
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currentOdd?.next = current;
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currentOdd = current;
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current = current.next;
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} else {
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currentEven?.next = current;
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currentEven = current;
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current = current.next;
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}
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i++;
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}
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currentOdd?.next = evenList.next;
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currentEven?.next = null;
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return oddList.next;
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}
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*/
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// func oddEvenList(head *ListNode) *ListNode {
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// if head == nil {
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// return head
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// }
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// oddNode := head
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// evenNode := head.Next
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// even := evenNode
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// for evenNode != nil && evenNode.Next != nil {
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// oddNode.Next = oddNode.Next.Next
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// evenNode.Next = evenNode.Next.Next
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// oddNode = oddNode.Next
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// evenNode = evenNode.Next
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// }
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// oddNode.Next = even
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// return head
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// }

‎OddEvenLinkedList/odd_even_linked_list.md

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# 🔥 Odd Even Linked List 🔥 || 3 Approaches || Simple Fast and Easy || with Explanation
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## Definition for singly-linked list
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```dart
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class ListNode {
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int val;
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ListNode? next;
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ListNode([this.val = 0, this.next]);
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}
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```
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## Solution - 1
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```dart
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class Solution {
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ListNode? oddEvenList(ListNode? head) {
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if (head != null) {
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ListNode? odd = head, even = head.next, evenHead = even;
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while (even != null && even.next != null) {
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odd?.next = odd.next?.next;
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even.next = even.next?.next;
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odd = odd?.next;
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even = even.next;
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}
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odd?.next = evenHead;
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}
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return head;
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}
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}
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```
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## Solution - 2
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```dart
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class Solution {
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ListNode? oddEvenList(ListNode? head) {
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if (head == null || head.next == null) {
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return head;
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}
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ListNode? evenList = ListNode(-1);
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ListNode? oddList = ListNode(-1);
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ListNode? currentEven = evenList;
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ListNode? currentOdd = oddList;
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ListNode? current = head;
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int i = 1;
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while (current != null) {
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if (i % 2 != 0) {
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currentOdd?.next = current;
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currentOdd = current;
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current = current.next;
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} else {
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currentEven?.next = current;
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currentEven = current;
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current = current.next;
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}
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i++;
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}
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currentOdd?.next = evenList.next;
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currentEven?.next = null;
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return oddList.next;
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}
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}
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```
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## Solution - 3
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```dart
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class Solution {
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ListNode? oddEvenList(ListNode? head) {
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ListNode? lastInEven = null;
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ListNode? temp = head;
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while (head != null && head.next != null) {
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ListNode? next = head.next;
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if (lastInEven == null) lastInEven = next;
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ListNode? temp = lastInEven?.next;
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if (temp == null) break;
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head.next = temp;
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lastInEven?.next = temp.next;
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temp.next = next;
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lastInEven = lastInEven?.next;
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head = head.next;
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if (lastInEven == null) break;
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}
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return temp;
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}
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}
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```

‎README.md

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@@ -159,6 +159,7 @@ This repo contain leetcode solution using DART and GO programming language. Most
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- [**451.** Sort Characters By Frequency](SortCharactersByFrequency/sort_characters_by_frequency.dart)
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- [**2256.** Minimum Average Difference](MinimumAverageDifference/minimum_average_difference.dart)
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- [**876.** Middle of the Linked List](MiddleOfTheLinkedList/middle_of_the_linked_list.dart)
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- [**328.** Odd Even Linked List](OddEvenLinkedList/odd_even_linked_list.dart)
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## Reach me via
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