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}.

基本思路：     用了一个很蠢很直接的方法，定位到中间节点，后半段入栈，然后把后半段与前半段拼接组成新链表

 

  1 /**
  2  * Definition for singly-linked list.
  3  * public class ListNode {
  4  *     int val;
  5  *     ListNode next;
  6  *     ListNode(int x) { val = x; }
  7  * }
  8  */
  9 public class Solution {
 10     public void reorderList(ListNode head) {
 11         
 12          
 13           if (head == null) 
 14             return;
 15           
 16           ListNode node = head.next;
 17           int len = 0;
 18           while(node != null){
 19               len ++;
 20               node = node.next;
 21           }
 22           
 23           System.out.println("len:"+len);
 24           
 25           if(len == 2){
 26               ListNode temp = head.next;
 27               
 28               head.next = temp.next;
 29               temp.next = null;
 30               head.next.next = temp;
 31               return;
 32           }else if(len == 1 || len == 0){
 33               return;
 34           }
 35           
 36           int isOdd = len % 2;
 37           if(isOdd == 1){
 38               int mid = len / 2 + 1;
 39               Stack<ListNode> stack = new Stack();
 40               
 41               int j = 1;
 42               ListNode midNode = head.next;
 43               while(j != mid){
 44                   j++;
 45                   midNode = midNode.next;
 46               }
 47               
 48               ListNode realMidNode = midNode;
 49               midNode = midNode.next;
 50               j++;
 51               stack.push(midNode);
 52               while(j != len){
 53                   j++;
 54                   midNode = midNode.next;
 55                   stack.push(midNode);
 56               }
 57               
 58               j = 1;
 59               
 60               ListNode leftTemp = head.next;
 61               ListNode rightTemp = stack.pop();
 62               rightTemp.next = leftTemp;
 63               head.next = rightTemp;
 64                ListNode tail = leftTemp;  
 65               j++;
 66               while(j < mid){
 67                   j++;
 68                   leftTemp = leftTemp.next;
 69                   rightTemp = stack.pop();
 70                   rightTemp.next = leftTemp;
 71                   tail.next = rightTemp;
 72                   tail = leftTemp;
 73               }
 74               
 75               tail.next = realMidNode;
 76               realMidNode.next = null;
 77               
 78           }else{
 79               int mid = len / 2 ;
 80               Stack<ListNode> stack = new Stack<>();
 81               
 82               int j = 1;
 83               ListNode midNode = head.next;
 84               while(j != mid){
 85                   j++;
 86                   midNode = midNode.next;
 87               }
 88               
 89               j++;
 90               midNode = midNode.next;
 91               stack.push(midNode);
 92               
 93               while(j != len){
 94                   j++;
 95                   midNode = midNode.next;
 96                   stack.push(midNode);
 97               }
 98               
 99               
100               
101               j=1;
102               
103               ListNode leftTemp = head.next;
104               ListNode rightTemp = stack.pop();
105               rightTemp.next = leftTemp;
106               head.next = rightTemp;
107                ListNode tail = leftTemp;  
108               
109               while(j != mid){
110                   j++;
111                   leftTemp = leftTemp.next;
112                   rightTemp = stack.pop();
113                   rightTemp.next = leftTemp;
114                   tail.next = rightTemp;
115                   tail = leftTemp;
116               }
117               tail.next = null;
118           }
119         
120     }
121 }
运行结果

 

 

posted @ 2016-11-25 09:59  music180  阅读(131)  评论(0)  编辑  收藏  举报