Middle-题目97：148. Sort List

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题目原文：
Sort a linked list in O(n log n) time using constant space complexity.
题目大意：
对一个单链表排序，要求时间复杂度O(nlogn),空间复杂度O(1)
题目分析：
参考discuss中的算法，使用快排，取pivot为第一个节点值。
源码：（language：java)

public class Solution {    public ListNode sortList(ListNode h){        if(h == null || h.next == null)            return h;        /*split into three list*/        ListNode fakesmall = new ListNode(0), small = fakesmall;        ListNode fakelarge = new ListNode(0), large = fakelarge;        ListNode fakeequal = new ListNode(0), equal = fakeequal;        ListNode cur = h; // pivot is h.        while(cur != null){            if(cur.val < h.val){                small.next = cur;                small = small.next;            }            else if(cur.val == h.val){                equal.next = cur;                equal = equal.next;            }            else{                large.next = cur;                large = large.next;            }            cur = cur.next;        }        // put an end.        small.next = equal.next = large.next = null;        // merge them and return . merge reusing below one. merge for quicksort should be simplified.         return merge(merge(sortList(fakesmall.next), sortList(fakelarge.next)),fakeequal.next) ;    }    private ListNode merge(ListNode h, ListNode m){        ListNode fake = new ListNode(0), cur = fake;        while(h != null && m != null){            if(h.val < m.val){                cur.next = h;                h = h.next;            }            else{                cur.next = m;                m = m.next;            }            cur = cur.next;        }        cur.next = (h == null ? m : h);        return fake.next;    }}

成绩：
8ms,beats 42.09%,众数8ms,31.36%
cmershen的碎碎念：
数组的快排中pivot是使用三者取中法，而链表不是随机存取的，使用三者取中法要遍历一次链表，反而不能加速。

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