143. Reorder List

先找中点.. 1 2 3 4 5 6 7 8
后半部分反转.. 1 2 3 4 8 7 6 5
分别链接。。 1 8 2 7 3 6 4 5

这种题思路不难，但是操作起来比较繁琐，每次做完一堆变量。

想不用这么多变量，就要用P1 P2这种名字上毫无意义的指针。。否则会误导善良，天真，无暇，无辜的群众。

public class Solution 
{
    public void reorderList(ListNode head) 
    {
        if(head == null || head.next == null || head.next.next == null) return;
        
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next != null && fast.next.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        

        ListNode temp = slow.next;
        slow.next = null;

        
        ListNode dummy = new ListNode(666);
        dummy.next = temp;
        ListNode prev = dummy;
        ListNode next = temp.next;
        

        
        temp.next = null;
        while(next != null)
        {
            prev = temp;
            temp = next;
            next = next.next;
            temp.next = prev;
        }
        
        
        ListNode temp1 = head;
        ListNode temp2 = temp;

        
        
        while(temp2 != null)
        {
            ListNode cur1 = temp1.next;
            temp1.next = temp2;
            temp1 = cur1;
            
            ListNode cur2 = temp2.next;
            temp2.next = temp1;
            temp2 = cur2;
        }
        
        
        
        head = dummy.next;
        
        
    }
}

Leetcode好像挂了，几个TEST跑了好久。