problem report: sort list

quick sort version 1:

　　quick sort 定义： https://en.wikipedia.org/wiki/Quicksort

　　quick sort 核心部分为partition（http://www.cnblogs.com/jiangchen/p/5398166.html），

　　　　　　　　　　　　取list中间点（http://www.cnblogs.com/jiangchen/p/5529704.html）

　　时间复杂度为o(nlgn)　　

 1 public class Solution {
 2     //recursive partition list to left,middle,right 3 parts, use concat function to conntet 3 parts together at end of function 
 3     public ListNode sortList(ListNode head) {  
 4         if (head == null || head.next == null) {
 5             return head;
 6         }
 7         ListNode middle = findMedian(head);
 8         
 9         ListNode leftDummy = new ListNode(0), leftTail = leftDummy;
10         ListNode rightDummy = new ListNode(0), rightTail = rightDummy;
11         ListNode middleDummy = new ListNode(0), middleTail = middleDummy;
12         
13         while (head != null) {
14             if (head.val < middle.val) {
15                 leftTail.next = head;
16                 leftTail = head;
17             } else if(head.val == middle.val) {
18                 middleTail.next = head;
19                 middleTail = head;
20             } else {
21                 rightTail.next = head;
22                 rightTail = head;
23             }
24             head = head.next;
25         }
26         leftTail.next = null;
27         middleTail.next = null;
28         rightTail.next = null;
29         
30         ListNode left = sortList(leftDummy.next);
31         ListNode right = sortList(rightDummy.next);
32         
33         return concat(left, middleDummy.next, right);
34     }
35     // connect 3 part together
36     private ListNode concat (ListNode left, ListNode mid, ListNode right) {
37         ListNode dummy = new ListNode(0), tail = dummy;
38         tail.next = left;
39         tail = getTail(tail);//left 本身可能是一个空串，所以3次getTail都用tail作为起始点，而不用left，mid，right
40         tail.next = mid;
41         tail = getTail(tail);
42         tail.next = right;
43         tail = getTail(tail);
44         return dummy.next;
45     }
46     //get end of current list 
47     private ListNode getTail (ListNode current) {
48         if (current == null) {
49             return current;
50         }
51         while (current.next != null) {
52             current = current.next;
53         }
54         return current;
55     }
56     
57     private ListNode findMedian(ListNode head) {
58         ListNode slow = head, fast = head.next;
59         while (fast != null && fast.next != null) {
60             slow = slow.next;
61             fast = fast.next.next;
62         }
63         return slow;
64     }
65 }

View Code

merge sort version 1:

　　merge sort 定义： https://en.wikipedia.org/wiki/Merge_sort

　　merge sort 核心部分为 recusive devide linked list to two part

　　　　　　　　　　　　　　merge a list

 1 public class Solution {            
 2     private ListNode findMiddle(ListNode head) {
 3         ListNode slow = head, fast = head.next;
 4         while (fast != null && fast.next != null) {
 5             fast = fast.next.next;
 6             slow = slow.next;
 7         }
 8         return slow;
 9     }    
10 
11     private ListNode merge(ListNode head1, ListNode head2) {
12         ListNode  tail,dummy = new ListNode(0);
13         tail = dummy;
14         while (head1 != null || head2 != null) {
15             if (head2 != null && (head1 == null || head1.val > head2.val)) {
16                 tail.next = head2;
17                 tail = head2;
18                 head2 = head2.next;
19             } else  {
20                 tail.next = head1;
21                 tail = head1;
22                 head1 = head1.next;
23             }
24         }
25        return dummy.next;
26     }
27 
28     public ListNode sortList(ListNode head) {
29         ListNode dummyHead1, dummyHead2;
30         if (head == null || head.next == null) {
31             return head;
32         }
33         ListNode head1, head2, mid;
34         mid = findMiddle(head);
35         head1 = head;
36         head2 = mid.next;
37         mid.next = null;
38        // System.out.println(head1.val);
39         //System.out.println(head2.val);
40 
41         head1 = sortList(head1);
42         head2 = sortList(head2);
43         
44         return merge(head1,head2);
45     }
46 }

View Code

 1 // version 1: Merge Sort
 2 public class Solution {            
 3     private ListNode findMiddle(ListNode head) {
 4         ListNode slow = head, fast = head.next;
 5         while (fast != null && fast.next != null) {
 6             fast = fast.next.next;
 7             slow = slow.next;
 8         }
 9         return slow;
10     }    
11 
12     private ListNode merge(ListNode head1, ListNode head2) {
13         ListNode dummy = new ListNode(0);
14         ListNode tail = dummy;
15         while (head1 != null && head2 != null) {
16             if (head1.val < head2.val) {
17                 tail.next = head1;
18                 head1 = head1.next;
19             } else {
20                 tail.next = head2;
21                 head2 = head2.next;
22             }
23             tail = tail.next;
24         }
25         if (head1 != null) {
26             tail.next = head1;
27         } else {
28             tail.next = head2;
29         }
30 
31         return dummy.next;
32     }
33 
34     public ListNode sortList(ListNode head) {
35         if (head == null || head.next == null) {
36             return head;
37         }
38 
39         ListNode mid = findMiddle(head);
40 
41         ListNode right = sortList(mid.next);
42         mid.next = null;
43         ListNode left = sortList(head);
44 
45         return merge(left, right);
46     }
47 }

View Code