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?…

Example
Given {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;
        }
    }