143. Reorder List

Posted 2021-02-15 markleebyr

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}.

class Solution {

    public void reorderList(ListNode head) {

        if (head == null || head.next == null)

            return;

        //找到链表的中点，如1 2 3 4 5 6 7 8，则p1是4

        ListNode p1 = head;

        ListNode p2 = head;

        while (p2.next != null && p2.next.next != null) {

            p1 = p1.next;

            p2 = p2.next.next;

        }

        

      //链表后半边逆序，5 6 7 8 变成 8 7 6 5 

        ListNode preMiddle = p1; //preMiddle 是4

        ListNode preCur = p1.next; //preCur 是5

        while (preCur.next != null) {

            ListNode cur = preCur.next; //cur是6

            preCur.next = cur.next; //5->7

            cur.next = preMiddle.next; //6->5

            preMiddle.next = cur; //4->6

        }

        //达到最后排序目的

        p1 = head;

        p2 = preMiddle.next;

        while (p1 != preMiddle) {

            preMiddle.next = p2.next; //另4->7

            p2.next = p1.next; //8->2

            p1.next = p2;//1->8

            p1 = p2.next;//让p1等于2

            p2 = preMiddle.next; //p2等于7，现在链表1 8 2 3 4 7 6 5

        }

        

    }

}