【LeetCode】Reorder List

다음으로 전송:http://www.programcreek.com/2013/12/in-place-reorder-a-singly-linked-list-in-java/
Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes' values.
For example, Given {1,2,3,4} , reorder it to {1,4,2,3} .
brute force 시간 초과,사고방식:

Break list in the middle to two lists (use fast & slow pointers)

Reverse the order of the second list

Merge two list back together

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
         if (head == null || head.next == null || head.next.next == null) {
			return ;
		}
		ListNode fast = head;
    	ListNode slow = head;
    	while (fast != null && fast.next != null) {
			fast = fast.next.next;
			slow = slow.next;
			
		}
    	ListNode head2 = null;
    	if (fast == null) {
			head2 = slow;
		}
    	else if (fast.next == null) {
			head2 = slow.next;
		}
    	slow = head;
    	while(slow.next != head2)
    		slow = slow.next;
    	slow.next = null;
    	
        head2 = reverse(head2);
        
        merge(head, head2);
    
    }
    public ListNode reverse(ListNode head)
    {
    	if (head == null || head.next == null) {
			return head;
		}
    	ListNode curListNode  = head.next;
    	ListNode preListNode = head;
    	while (curListNode.next != null) {
    		ListNode tmp = curListNode.next;
			curListNode.next = preListNode;
			preListNode = curListNode;
			curListNode = tmp;
		}
    	curListNode.next = preListNode;
    	head.next = null;
    	head = curListNode;
    	return head;
    }
    public void merge(ListNode head1, ListNode head2)
    {
    	ListNode p1 = head1;
    	ListNode p2 = head2;
    	while (p1.next != null && p2.next != null) {
			ListNode p1_next = p1.next;
			ListNode p2_next = p2.next;
			p1.next = p2;
			p2.next = p1_next;
			p1 = p1_next;
			p2 = p2_next;
			
		}
        ListNode p1_next = p1.next;
    	p1.next = p2;
    	if (p1_next != null) {
			p2.next = p1_next;
		}
    	
    }
}