LeetCode: Sort List 解题报告

 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         // Nodes should be more than 2.
15         if (head == null || head.next == null) {
16             return head;
17         }
18         
19         // get the mid node.
20         ListNode midPre = getMidPre(head);
21         
22         // Cut the two list.
23         ListNode right = midPre.next;
24         midPre.next = null;
25         
26         // Sort the left side and the right side.
27         ListNode left = sortList(head);
28         right = sortList(right);
29         
30         // Merge the two sides together.
31         return merge(left, right);
32     }
33     
34     // get the pre node before mid.
35     public ListNode getMidPre1(ListNode head) {
36         ListNode slow = head;
37         ListNode fast = head;
38         
39         while (fast != null && fast.next != null && fast.next.next != null) {
40             slow = slow.next;
41             fast = fast.next.next;
42         }
43         
44         return slow;
45     }
46     
47     // get the pre node before mid.
48     public ListNode getMidPre(ListNode head) {
49         ListNode slow = head;
50         
51         // fast提前一点儿。这样就可以得到前一个节点喽。
52         ListNode fast = head.next;
53         
54         while (fast != null && fast.next != null) {
55             slow = slow.next;
56             fast = fast.next.next;
57         }
58         
59         return slow;
60     }
61     
62     public ListNode merge(ListNode head1, ListNode head2) {
63         ListNode dummy = new ListNode(0);
64         ListNode cur = dummy;
65         
66         while (head1 != null && head2 != null) {
67             if (head1.val < head2.val) {
68                 cur.next = head1;
69                 head1 = head1.next;
70             } else {
71                 cur.next = head2;
72                 head2 = head2.next;
73             }
74             
75             cur = cur.next;
76         }
77         
78         if (head1 != null) {
79             cur.next = head1;
80         } else {
81             cur.next = head2;
82         }
83         
84         return dummy.next;
85     }
86 }