【Leetcode】Reorder List

题目链接：https://leetcode.com/problems/reorder-list/

题目：

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：

划分后半段链表，找到它的表头。然后将它反转，再和前半段链表交叉合并。其中划分链表、反转部分代码在Palindrome Linked List 中已有实现，

只需添加交叉合并操作就可以了。

算法：

	public void reorderList(ListNode head) {
		if (head == null || head.next == null) {
			return;
		}
		int i = getListLength(head) / 2;
		ListNode p = head, l2 = null;
		while (i-- > 0)
			p = p.next;
		l2 = p.next; //此时l2指向后半段链表表头
		p.next = null;
		
		l2 = reverseList(l2);// 反转后半段链表

		p = head;
		while (l2 != null) {// 将两链表交叉合并，在p~t之间插入l2结点，其中t2指向l2后面结点
			ListNode t = p.next;
			ListNode t2 = l2.next;
			p.next = l2;
			l2.next = t;
			l2 = t2;
			p = t;
		}
	}

	public int getListLength(ListNode head) {
		int length = 0;
		ListNode p = head;
		while (p != null) {
			length++;
			p = p.next;
		}
		return length;
	}

	public ListNode reverseList(ListNode head) {
		ListNode p = head, q, t;
		if (p == null)
			return head;
		q = p.next;
		while (q != null) {
			t = q.next;// 保存当前要处理结点后面的一个结点
			q.next = p;
			p = q; // p是新链表头结点
			q = t;
		}
		head.next = null;// 原头结点变成尾节点
		head = p;
		return head;
	}