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.
一个指针记录奇数的结尾，一个指针记录偶数的开头，一个指针记录偶数的末尾。
奇数的结尾的next，指向偶数的开头，偶数的结尾的next是下一个奇数，奇数的next是下一个偶数
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode oddEvenList(ListNode head) {
if (head == null || head.next == null || head.next.next == null) {
return head;
}
ListNode p = head.next;
ListNode oddTail = head;
ListNode evenHead = head.next;
ListNode preEven;
while (true) {
if (p.next == null) {
break;
}
preEven = p;
p = p.next;
oddTail.next = p;
ListNode nextEven = p.next;
oddTail = p;
p.next = evenHead;
preEven.next = nextEven;
p = nextEven;
if (p == null) {
break;
}
}
return head;
}
}