Leetcode: Insertion Sort List

Sort a linked list using insertion sort.

 我原本的想法是用额外的空间拷贝每一个节点，建立了一个新的sorted的LinkedList, 后来看到别人的做法不用建立新的LinkedList,  直接以原有List上的节点组成新的sorted的LinkedList。我之前想这样做会扰乱遍历的顺序，但是其实sorted的list和unsorted list是完全分开互不影响的。先还是给sorted list建立一个dummy node做前置节点， 每次取unsorted list里面的一个节点，记录下下一跳位置，然后把这个点插入到sorted list对应位置上（pre.next指到null或者pre.next。val更大）。

Insert Sort的做法相当于是每次从原来的List里面删除头节点，再把这个头节点插入到新的List里相应的位置。这个新List全由原来的节点组成，只是变换了顺序。

时间复杂度是插入排序算法的O(n^2)，空间复杂度是O(1)。

第二遍做法：

 1 public class Solution {

 2     public ListNode insertionSortList(ListNode head) {

 3         ListNode dummy = new ListNode(-1);

 4         ListNode cursor = dummy;

 5         while (head != null) {

 6             ListNode next = head.next;

 7             while (cursor.next!=null && cursor.next.val<=head.val) {

 8                 cursor = cursor.next;

 9             }

10             head.next = cursor.next;

11             cursor.next = head;

12             head = next;

13             cursor = dummy;

14         }

15         return dummy.next;

16     }

17 }

第一遍做法：

 1 /**

 2  * Definition for singly-linked list.

 3  * public 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 insertionSortList(ListNode head) {

14         if(head == null)

15             return null;

16         ListNode helper = new ListNode(0);

17         ListNode pre = helper;

18         ListNode cur = head;

19         while(cur!=null)

20         {

21             ListNode next = cur.next;

22             pre = helper;

23             while(pre.next!=null && pre.next.val<=cur.val)

24             {

25                 pre = pre.next;

26             }

27             cur.next = pre.next;

28             pre.next = cur;

29             cur = next;

30         }

31         return helper.next;

32       }

33 }

上面程序注释如下：

 1 public ListNode insertionSortList(ListNode head) {

 2     if(head == null)

 3         return null;

 4     ListNode helper = new ListNode(0);//helper is the dummy head for the result sorted LinkedList

 5     ListNode pre = helper; //pre is the pointer to find the suitable place to plug in the current node

 6     ListNode cur = head; //cur is current node, the node to be plugged in the sorted list

 7     while(cur!=null)

 8     {

 9         ListNode next = cur.next; //keep a record of the current node's next

10         pre = helper;  //after one search, put pre back to its original place

11         while(pre.next!=null && pre.next.val<=cur.val) //use pre to traverse the sorted list to find the suitable place to plug in

12         {

13             pre = pre.next;

14         }

15         cur.next = pre.next;// plug in current node to the right place

16         pre.next = cur;

17         cur = next; //go on to deal with the next node in the unsorted list

18     }

19     return helper.next;

20 }