143. Reorder List

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}.

一刷
题解：
先将median的之后的节点reverse, 然后再逐一连起来

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null) return;
        
        ListNode p1 = head;
        ListNode p2 = head;
        while(p2.next!=null && p2.next.next!=null){
            p1 = p1.next;
            p2 = p2.next.next;
        }
        //then p1 indicate the medium
        //reverse after the medium
        // 1->2->3->4->5->6 to 1->2->3->6->5->4
        ListNode preMid = p1;
        ListNode preCur = p1.next;
        
        //preMid.next indicate the head, cur connect with the preMid.next
        while(preCur.next!=null){
            ListNode cur = preCur.next;
            preCur.next = cur.next;
            cur.next = preMid.next;//connect the head
            preMid.next = cur;//change the head
        }
        
        
        p1 = head;
        p2 = preMid.next;
        while(p1!=preMid){
            preMid.next = p2.next;//change the head
            p2.next = p1.next;
            p1.next = p2;
            p1 = p2.next;
            p2 = preMid.next;
        }
        
    }
}

二刷
思路同一刷
不过要注意快慢指针的停止位置： while(fast.next!=null && fas t.next.next!=null)
这样保证从slow的后面开始reverse

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        //first reverse half and linked each other from the both head
        //find the medium
        if(head == null || head.next==null) return;
        ListNode fast = head, slow = head;
        while(fast.next!=null && fast.next.next!=null){
            slow = slow.next;
            fast = fast.next.next;
        }
        
        //reverse after the medium
        ListNode prev = null, cur = slow.next;
        
        while(cur!=null){
            ListNode next = cur.next;
            cur.next = prev;
            slow.next = cur;
            prev = cur;
            cur = next;
        }
        
        ListNode first = head, second = slow.next;
        slow.next = null;
        while(first!=null && second!=null){
            ListNode fNext = first.next;
            ListNode sNext = second.next;
            first.next = second;
            second.next = fNext;
            first = fNext;
            second = sNext;
        }
    }
}