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

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。
下面的源码，用到的技巧有快慢指针，获取链表的中间指针。
链表存储的归并排序 时间复杂度O（nlogn）空间复杂度 O（1）

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {        //快速排序，空间复杂度为O(n)   /* public ListNode sortList(ListNode head) {        quickSort(head,null);        return head;    }        public static void quickSort(ListNode head, ListNode end){        if (head != end){            ListNode partion = partion(head);            quickSort(head,partion);            quickSort(partion.next,end);        }    }        public static ListNode partion(ListNode head){        ListNode slow = head;        ListNode fast = head.next;                while (fast != null){            if (fast.val < head.val){                slow = slow.next;                fast.val = slow.val ^ fast.val ^(slow.val = fast.val);            }            fast = fast.next;        }        slow.val = head.val ^ slow.val ^ (head.val = slow.val);        return slow;    }*/        //归并排序    public ListNode sortList(ListNode head) {        if (head == null || head.next == null){            return head;        }                ListNode fast = head;        ListNode 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;    }}