Sort List
Sort a linked list in O(n log n) time using constant space complexity.
这道题我想过用直接插入，超时了。百度了一下，用的是归并排序，排序以前用的都是数组，这次用链表，不太习惯
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1 /**
2 * Definition for singly-linked list.
3 * class ListNode {
4 * int val;
5 * ListNode next;
6 * ListNode(int x) {
7 * val = x;
8 * next = null;
9 * }
10 * }
11 */
12 public class Solution {
13 public ListNode sortList(ListNode head) {
14 if(null == head || null == head.next)
15 return head; //空链表和只有一个节点的链表直接返回head
16
17 ListNode middle = getMiddle(head); //得到中点节点
18 ListNode next = middle.next;
19 middle.next = null;
20
21 return merge(sortList(head), sortList(next));
22 }
23
24 /**
25 * 获取链表的中间位置节点，一个指针走一步，另一个指针走两步
26 * @param head
27 * @return
28 */
29 public ListNode getMiddle(ListNode head){
30 ListNode fast = head;
31 ListNode slow = head;
32
33 while(null != fast.next && null != fast.next.next){
34 fast = fast.next.next;
35 slow = slow.next;
36 }
37
38 return slow;
39 }
40
41 /**
42 * 合并两个有序链表
43 * @param a
44 * @param b
45 */
46 public ListNode merge(ListNode a, ListNode b){
47 ListNode dummyHead = new ListNode(Integer.MIN_VALUE);
48 ListNode cur = dummyHead;
49 while(null != a && null != b){ //开始合并两个链表，放到dummyHead为头结点的链表中
50 if(a.val <= b.val){
51 cur.next = a;
52 a = a.next;
53 }
54 else{
55 cur.next = b;
56 b = b.next;
57 }
58 cur = cur.next;
59 }//while
60 cur.next = a != null ? a : b;
61
62 return dummyHead.next;
63 }
64 }
