对链表排序

一、要求：

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

二、分析：

使用o(nlogn)的时间复杂度和常量空间复杂度，对链表排序，只能使用归并排序。归并排序是将两个或两个以上的有序链表合并成一个新的链表。常见的是二路归并排序算法，思想是将数组或链表中前后相邻的两个有序序列归并为一个有序序列，时间复杂度为o(nlogn)，需要等数量的辅助空间。

三、代码实现：

题目要求对链表进行排序，时间复杂度为o(nlogn)，空间复杂度为常亮级的，按照要求，代码如下：

<span style="font-size:18px;"></span>

<pre name="code" class="java"><span style="font-size:14px;">/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode sortList(ListNode head) {                if(head == null || head.next == null){             return head;         }         ListNode fast = head, slow = head;         while(fast != null && fast.next != null && fast.next.next != null){             fast = fast.next.next;             slow = slow.next;         }         fast = slow;         slow = slow.next;         fast.next = null;         fast = sortList(head);         slow = sortList(slow);         return merge(fast, slow);      }      private ListNode merge(ListNode sub1, ListNode sub2){          if(sub1== null){              return sub2;          }          if(sub2 == null){              return sub1;          }          ListNode head;          if(sub1.val < sub2.val){              head = sub1;              sub1 = sub1.next;          }else{              head = sub2;              sub2 = sub2.next;          }          ListNode p = head;          while(sub1 != null && sub2 != null){              if(sub1.val < sub2.val){                  p.next = sub1;                  sub1 = sub1.next;              }else{                  p.next = sub2;                  sub2 = sub2.next;              }              p = p.next;          }          if(sub1 != null){              p.next = sub1;          }          if(sub2 != null){              p.next = sub2;          }          return head;     }}</span>