# Day 16 of 31-Day May LeetCode Challenge

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

`/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* oddEvenList(ListNode* head) {                if(head==NULL||head->next==NULL) return head;        //         //method 1 : O(n) time & O(n) space  -> 35 %//         ListNode *odd = new ListNode(-1);   //added a dummy node in the beginning//         ListNode *even = new ListNode(-1);  //added a dummy node in the beginning        //         ListNode *myHead = odd;        //         //creating odd linked list//         ListNode *temp = head;//         while(true){//             odd->next = new ListNode(temp->val);//             odd = odd->next;//             if(temp->next!=NULL&&temp->next->next!=NULL){//                 temp = temp->next->next;//             }else break;//         }//         //creating even linked list//         ListNode *evenHead = even;//         ListNode *temp2 = head->next;//         while(true){//             even->next = new ListNode(temp2->val);//             even = even->next;//             if(temp2->next!=NULL&&temp2->next->next!=NULL){//                 temp2 = temp2->next->next;//             }else break;//         }//         //linking odd & even list//         odd->next = evenHead->next;        //         return myHead->next;                        //method 2 : O(n) time & O(1) space -> 95 %        //hint 1 : we need to make odd & even linked list SIMULTANOUSLY        //hint 2 : we need 4 pointers        ListNode *odd = head;        ListNode *even = head->next;        ListNode *evenHead = even;        ListNode *oddHead = odd;                while(even!=NULL&&even->next!=NULL){            odd->next = even->next;            even->next = even->next->next;                        odd = odd->next;            even = even->next;        }        odd->next = evenHead;                return oddHead;            }};`