【Leetcode】Sort List JAVA实现

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1、题：

Sort a linked list in O(n log n) time using constant space complexity.

2、分析

该题主要考查了链接上的合并排序算法。

3、代码实现

package com.edu.leetcode;import com.edu.leetcode.ListNode;public class SortList {/** * @param args */public ListNode Merge(ListNode first,ListNode second){   //合并两个有序链表，并返回新链表的投结点ListNode rear;ListNode head;rear=head=new ListNode(-1);while(first!=null&&second!=null){if(first.val<=second.val){rear.next=first;rear=first;first=first.next;}else{rear.next=second;rear=second;second=second.next;}}if(first!=null)rear.next=first;elserear.next=second;return head.next;}public ListNode sortList(ListNode head){     /* * 实现链表的合并排序：1、将链表划分成基本相等的两个链表 * 2、递归将这两个链接继续划分，直到链表的长度为0或者1为止 * 3、调用Merge（）将链接进行合并 */if(head==null||head.next==null)return head;ListNode mid =head;ListNode pos =mid.next;while(pos!=null){pos=pos.next;if(pos!=null){pos=pos.next;mid=mid.next;}}ListNode q=sortList(mid.next);mid.next=null;return Merge(sortList(head), q);}public static void main(String[] args) {// TODO Auto-generated method stubListNode first1 = new ListNode(0);ListNode rear1 =first1;for(int i=9;i>=1;i--){ListNode q= new ListNode(i);rear1.next=q;rear1=q;}ListNode q=first1;while(q!=null){System.out.print(q.val+ ",");q=q.next;}System.out.println();SortList sl = new SortList();sl.sortList(first1);ListNode p=first1;while(p!=null){System.out.print(p.val+ ",");p=p.next;}System.out.println();}}

4.疑问

这里的问题主要是将上面的SortList（）方法分成两步实现：Divide()和

MergeSort（）两部分。个人觉得应该是一样的。但运行结果确实不同，如果大神浏

览，请指点一下小弟。。

class ListNode{int val;ListNode next=null;public ListNode(){}public ListNode(int val){this.val=val;}}public class SortList {public ListNode Merge(ListNode first,ListNode second){ListNode rear;ListNode head;rear=head=new ListNode();while(first!=null&&second!=null){if(first.val<=second.val){rear.next=first;rear=first;first=first.next;}else{rear.next=second;rear=second;second=second.next;}}if(first!=null)rear.next=first;elserear.next=second;return head.next;}public ListNode Divide(ListNode first){   <span style="line-height: 1.5; font-family: 'Courier New'; white-space: pre-wrap;">//将一个链表划分成两个基本相等的子链表</span>if(first==null)return null;ListNode mid =first;ListNode pos =mid.next;while(pos!=null){pos=pos.next;if(pos!=null){pos=pos.next;mid=mid.next;}}ListNode q=mid.next;mid.next=null;return q;}public void MergeSort(ListNode first){if(first!=null&&first.next!=null){ListNode second =Divide(first);MergeSort(first);MergeSort(second);Merge(first, second);}}public static void main(String[] args) {// TODO Auto-generated method stubListNode first1 = new ListNode(0);ListNode rear1 =first1;for(int i=9;i>=1;i--){ListNode q= new ListNode(i);rear1.next=q;rear1=q;}ListNode q=first1;while(q!=null){System.out.print(q.val+ ",");q=q.next;}System.out.println();SortList sl = new SortList();sl.MergeSort(first1);ListNode p=first1;while(p!=null){System.out.print(p.val+ ",");p=p.next;}System.out.println();}}

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