Reorder List @LeetCode

思路就是：

1 用快慢指针找到中间节点

2 翻转中间节点后一个元素到最后一个元素区间的所有元素

3 断开前半段和翻转后的后半段元素

4 把前半段和翻转后的后半段元素以交叉的方式合并起来

5 特殊处理输入为空，只有一个元素和只有两个元素的corner case，就是多加几个if...return

感想：可以看出翻转链表实在是非常重要，是做很多题目的基础。还有merge的思想也很重要！

 

package Level4;



import Utility.ListNode;



/**

 *

 * Reorder List 

 *

 * 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}.

 *

 */

public class S143 {



	public static void main(String[] args) {

		int[] list = {1};

		ListNode head = ListNode.create(list);

		reorderList(head);

	}



	public static void reorderList(ListNode head) {

		ListNode fast = head;

		ListNode slow = head;

		// 找到中间节点

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

			fast = fast.next.next;

			slow = slow.next;

		}

		

//System.out.println(slow.val);

		ListNode preReverse = slow;	// preReverse不用翻转，因为它永远在最后一个

		if(preReverse == null){

			return;

		}

		

		// 翻转后半段

		ListNode reHead = preReverse.next;

		if(reHead == null){

			return;

		}

		ListNode preCur = reHead.next;

		ListNode cur = reHead.next;

		reHead.next = null;

		while(cur != null){

			cur = cur.next;

			preCur.next = reHead;

			reHead = preCur;

			preCur = cur;

		}

		preReverse.next = reHead;

//head.print();

		

		// 交叉合并两个链表

		preReverse.next = null;		// 断开前半段和翻转后的后半段元素

		cur = head;

		while(reHead != null && cur!=null){

			ListNode tmp = cur.next;

			cur.next = reHead;

			reHead = reHead.next;

			cur.next.next = tmp;

			cur = tmp;

			tmp = cur.next;

		}

//head.print();

	}



}