leetcode--Reorder List

143.题目

Given a singly linked list L: L0?L1?…?Ln-1?Ln,reorder it to: L0?Ln?L1?Ln-1?L2?Ln-2?…ExampleGiven {1,2,3,4}, reorder it to {1,4,2,3}.

第一反应写的代码

class Solution {    public void reorderList(ListNode head) {        if(head == null) return;        if(head.next == null) return;        ListNode pre = head,preLast = head.next;        if(preLast.next == null) return;        while(preLast.next != null){            pre = pre.next;            preLast = pre.next;        }        ListNode next = head.next;        head.next = preLast;        preLast.next = next;        pre.next = null;        reorderList(next);        return;    }}

总共13组测试用例，跑过了12组，最后一组特别大的显示Time Limit Exceeded
改进……
1.可以找list的中心，从中心分为两个list；
2.对后半部分的实现从后指向前；
3.剩下两个list插空填充就行了。
下面是实现：

public void reorderList(ListNode head) {        if(head == null || head.next == null) return;        ///find the middle        ListNode p1 = head;        ListNode p2 = head;        while(p2.next != null && p2.next.next != null){            p1 = p1.next;            p2 = p2.next.next;        }        if(p2.next != null)            p2 = p2.next;        //p1 in the middle of the list        //p2 last in the list        //reverse the latter half of the list 1->2->3->4->5->6 to 1->2->3<-4<-5<-6        ListNode preCurrent = p1;        ListNode current = p1.next;        //if(p1 != head)        p1.next = null;        while(p1 != head && current != null){            ListNode k1 = current.next;            current.next = preCurrent;            preCurrent = current;            current = k1;        }        //merge two list 1->2->3 6->5->4->3        ListNode tmp1 = head;        ListNode tmp2 = p2;        while(tmp1 != tmp2 && tmp1 != null && tmp2 != null){            ListNode k1 = tmp1.next;            ListNode k2 = tmp2.next;            tmp1.next = tmp2;            tmp2.next = k1;            tmp1 = k1;            tmp2 = k2;        }    }