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}.

思路：

1. 找出中间节点

2. 把中间节点之后的后半部分链表反序

3. 把前半部分链表及后半部分链表合并

4. 把中间节点 指向 null

/**
 * 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) {
       int count = 0;
       ListNode midPoint = head;
       ListNode fastRunner = head;
       // while 结束时 fastRunner指向的是最后一个点 mid指向的是中点，如果总数是偶数，指向的是晓得值
       while(fastRunner != null && fastRunner.next != null){
           fastRunner = fastRunner.next.next;
           midPoint = midPoint.next;
           count++;
       }
        if(count < 1 || midPoint.next == null)
            return;
        
        fastRunner = midPoint.next;
        ListNode runner = midPoint;
        while(fastRunner != null){
            ListNode tmp = fastRunner.next;
            fastRunner.next = runner;
            runner = fastRunner;
            fastRunner = tmp;
        }
        
        ListNode frontRunner = head;
        ListNode backRunner = runner;
        while(frontRunner != midPoint && backRunner != midPoint){
            ListNode tmp = backRunner.next;
            backRunner.next = frontRunner.next;
            frontRunner.next = backRunner;
            backRunner = tmp;
            frontRunner = frontRunner.next.next;
        }
        midPoint.next = null;
    }
    
}