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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.
/**
* 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) return head;
ListNode odd = head;
ListNode even = head.next;
ListNode even_head = even;
while (odd.next != null && even.next != null) {
odd.next = even.next;
odd = even.next;
even.next = odd.next;
even = odd.next;
}
odd.next = even_head;
return head;
}
}
两位两位向后跑，奇数链表的尾指针会被Null替代；偶数链表的尾指针本身就是Null。
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原文地址：http://www.cnblogs.com/Raymond-Yang/p/5166704.html
