# Leetcode 328 - Odd even linked list

`Note:`

- Think about doing the same thing to an array in O(n) speed and O(1) extra space. It’s kinda hard.
- What makes linked list easier? You can
`ignore it's position`

in the input array and just assign their`next`

. - Use
`two pointers`

, one to odd, one to even. - Let
`odd pointer`

point to next`odd position`

. - Let
`even pointer`

point to next`even position`

. - Move them to next.
- Concat
`even pointer`

to`odd pointer`

to connect them together. - Return the head.

`Question:`

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.

`Example:`

1 | Input: head = [2,1,3,5,6,4,7] |

`Code:`

1 | /** |