https://leetcode.com/problems/odd-even-linked-list/description/
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.
Example:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.
Note:
The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on ...
1 public ListNode oddEvenList(ListNode head) {
2 if (head == null || head.next == null) return head;
3 ListNode oddCurr = head;
4 ListNode evenHead = head.next ;
5 ListNode evenCurr = head.next;
6 /*
7 1->2->3->4->5->NULL
8 co ce
9 eh
10 1->3 2->4
11 ->
12 * */
13 int count = 0;
14 /*
15 even is always one node ahead so:
16 evenCurr != null == oddCurr.next != null
17 evenCurr.next != null == oddCurr.next.next != null
18
19 重要！when evenCurr is at 4, evenCurr.next = 5 != null
20 */
21 while (evenCurr != null && evenCurr.next != null){
22 oddCurr.next = oddCurr.next.next ;
23 oddCurr = oddCurr.next;
24 //4->5.next=null 4->null
25 evenCurr.next = evenCurr.next.next ;
26 evenCurr = evenHead.next;
27 }
28 //奇数结尾指向偶数的头
29 oddCurr.next = evenHead ;
30 return head;
31 }
