Medium
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 ...
这道题一开始固定head和head.next(even),用odd和curtEven来移动指针。题意表面上是说把偶数项都移后，但实际上完全可以转换思路为将奇数节点和偶数节点分离。比如这里分别在奇数项链里删掉偶数项链，同时也把even节点每隔一个连接起来，就成了全是even的链表。知道odd不能再用odd.next = odd.next.next删节点，说明这时候odd节点已经遍历完了，只需要把even节点接到后面就可以了。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode oddEvenList(ListNode head) {
if (head == null){
return head;
}
ListNode odd = head;
ListNode even = head.next;
ListNode curtEven = even;
while (odd.next != null && odd.next.next != null && curtEven != null){
odd.next = odd.next.next;
odd = odd.next;
curtEven.next = odd.next;
curtEven = curtEven.next;
}
odd.next = even;
return head;
}
}
