In this **Leetcode Odd-Even Linked List problem solution**, You are 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.

The first node is considered odd, and the second node is even, and so on. Note that the relative order inside both the even and odd groups should remain as it was in the input. You must solve the problem in O(1) extra space complexity and O(n) time complexity.

## Problem solution in Python.

class Solution(object): def oddEvenList(self, head): def helper(odd, even, node, level): if not node: return (odd, even) if level % 2 == 1: odd.next = node node = node.next odd = odd.next odd.next = None else: even.next = node node = node.next even = even.next even.next = None return helper(odd, even, node, level + 1) dummyOdd = ListNode(0) dummyEven = ListNode(0) odd, even = helper(dummyOdd, dummyEven, head, 1) odd.next = dummyEven.next return dummyOdd.next

## Problem solution in Java.

public ListNode oddEvenList(ListNode head) { if (head == null) return head; ListNode evenHead = head.next, oddWalker = head, evenWalker = evenHead; while(evenWalker != null && evenWalker.next != null) { oddWalker.next = oddWalker.next.next; evenWalker.next = evenWalker.next.next; oddWalker = oddWalker.next; evenWalker = evenWalker.next; } oddWalker.next = evenHead; return head; }

## Problem solution in C++.

ListNode* oddEvenList(ListNode* head) { ListNode **odd = &head, *even_head = NULL, **even = &even_head; while (*odd) { odd = &((*odd)->next); if (*even = *odd) { *odd = (*odd)->next; even = &((*even)->next); } } *odd = even_head; return head; }

## Problem solution in C.

struct ListNode* oddEvenList(struct ListNode* head) { if(head==NULL || head->next==NULL) return head; struct ListNode *p=head; struct ListNode *q=p->next; while(q!=NULL && q->next!=NULL) { struct ListNode *tmp=q->next; q->next = q->next->next; tmp->next=p->next; p->next=tmp; p=p->next; q=q->next; } return head; }

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