leetcode OddEvenLinkedList
z
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
import utils.ListNode;

public class OddEvenLinkedList {
/**
* 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 1:
*
* Input: 1->2->3->4->5->NULL
* Output: 1->3->5->2->4->NULL
* Example 2:
*
* Input: 2->1->3->5->6->4->7->NULL
* Output: 2->3->6->7->1->5->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 ...
*/
public ListNode oddEvenList(ListNode head) {
if (head == null || head.next == null) return head;
ListNode dump = new ListNode(0);
dump.next = head;
ListNode odd = head;
ListNode even = head.next;
ListNode p = even.next;
while(p!=null){
//insert(odd.next, p);
even.next = p.next;
ListNode temp = odd.next;
odd.next = p;
p.next = temp;

odd = odd.next;
even = even.next;
if (even == null) break;
p = even.next;
}
return dump.next;
}
public ListNode oddEvenListII(ListNode head) {
if (head == null || head.next == null) return head;
ListNode headOdd = head;
ListNode headEven = head.next;
ListNode tailOdd = head;
ListNode taillEven = head.next;
while (taillEven.next != null) {
ListNode t = taillEven.next;
tailOdd.next = t;
taillEven.next = t.next;
t.next = headEven;
taillEven = taillEven.next;
tailOdd = tailOdd.next;
}
return headOdd;
}
}