每日AC--leetcode--recordList

由于最近大脑短路的十分厉害， 表示很无奈

题目描述

Given a singly linked list L: L 0→L 1→…→L n-1→L n,

reorder it to: L 0→L n →L 1→L n-1→L 2→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}.

AC代码：

public class ListNode {
public int val;
public ListNode next = null;

ListNode(int val) {
this.val = val;
}
}

/**
* 类说明
*
* <pre>
* Modify Information:
* Author        Date          Description
* ============ =========== ============================
* DELL          2017年7月25日    Create this file
* </pre>
*
*/

public class LeetCodeRecordList {
/**
*
Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→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}.
题目的意思和明显 0 1  2  3
n n-1 n-2
* @param head
*/
public void reorderList(ListNode head) {
// 链表的结构
// node1->node2->node3->null
ListNode rear = head;
ListNode tmp ; //构建新的链表 尾插法建立链表
while(head != null && head.next != null){
// 先定位到尾节点 前一个  例如这里算 node2
while(rear.next.next!= null){
rear = rear.next;
}

tmp =  rear.next ; //取出尾节点后一个 比如  node3
rear.next = null; // 原队列node2的后一个置为null
tmp.next = head.next;
head.next = tmp;

head = tmp.next; // 下一次循环从 node1 开始
rear = tmp.next;
}

}

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
ListNode node1 = new ListNode(1);
ListNode node2 = new ListNode(2);
ListNode node3 = new ListNode(3);
ListNode node4 = new ListNode(4);
node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = null;
new LeetCodeRecordList().reorderList(node1);

while(node1 !=null){
System.out.println(node1.val);
node1 = node1.next;
}
}

}