328. Odd Even Linked List
Medium
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 = [1,2,3,4,5]
Output: [1,3,5,2,4]
Example 2:
Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]
Constraints:
- The number of nodes in the linked list is in the range
[0, 104]. -106 <= Node.val <= 106
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func oddEvenList(head *ListNode) *ListNode {
if head == nil { return nil }
odd := &ListNode{}
oddH := odd
even := &ListNode{}
evenH := even
flag := 1
for head != nil {
if flag == 1 {
odd.Next = head
odd = odd.Next
flag = 0
} else {
even.Next = head
even = even.Next
flag = 1
}
head = head.Next
}
even.Next = nil
odd.Next = evenH.Next
return oddH.Next
}