[LeetCode]Reorder List

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n, reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

思考：[微软原题点击]。O(n²)方法这里就不贴了。因为每次要插入的结点都是尾结点，从头结点开始寻找时间都花费在查询上。题意说只可以改变.next，这算是一个提示吧。我们可以翻转待插入链表结点，这样每次查询尾结点时间为O(1)，极大减小时间复杂度。翻转链表，合并链表，非常棒的一道题目。

class Solution {

public:

	ListNode *Reverse(ListNode *head)

	{

		if(!head||!head->next) return head;

		ListNode *p=head;

		ListNode *q=p->next;

		while(q)

		{

			p->next=q->next;

			q->next=head;

			head=q;

			q=p->next;

		}

		return head;

	}

	ListNode *Merge(ListNode *head1,ListNode *head2)

	{

		ListNode *p=head1;

		ListNode *q=head2;

		ListNode *pN,*qN;

		while(q)

		{

			pN=p->next;

			qN=q->next;

			p->next=q;

			q->next=pN;

			p=pN;

			q=qN;

		}

		return head1;

	}

    void reorderList(ListNode *head) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

		if(!head||!head->next||!head->next->next) return;

		ListNode *p;

		p=head;

		int num=0;

		while(p)

		{

			p=p->next;

			num++;

		}

		if(num%2) num=num/2;

		else num=num/2-1;

		p=head;

		while(num--)

			p=p->next;

		ListNode *head2=p->next;

		p->next=NULL;

		head2=Reverse(head2);

		head=Merge(head,head2);

    }

};