Insertion Sort List

这题一开始想新建一个list，结果MLE了，后来想了想不用新建，insertion的概念理解好就行，具体编程部分不难

 1 /**

 2  * Definition for singly-linked list.

 3  * struct ListNode {

 4  *     int val;

 5  *     ListNode *next;

 6  *     ListNode(int x) : val(x), next(NULL) {}

 7  * };

 8  */

 9 class Solution {

10 public:

11     ListNode *insertionSortList(ListNode *head) {

12         if (!head) return NULL;

13         ListNode *hpre = new ListNode(0);

14         hpre->next = head;

15         ListNode *pre = head;

16         head = head->next;

17         while (head) {

18             if (head->val > pre->val) {

19                 head = head->next;

20                 pre = pre->next;

21             }

22             else {

23                 pre->next = head->next;

24                 ListNode *tmp = hpre;

25                 while (head->val > tmp->next->val) tmp = tmp->next;

26                 head->next = tmp->next;

27                 tmp->next = head;

28                 head = pre->next;

29             }

30         }

31         return hpre->next;

32     }

33 };

 C#

 1 /**

 2  * Definition for singly-linked list.

 3  * public class ListNode {

 4  *     public int val;

 5  *     public ListNode next;

 6  *     public ListNode(int x) { val = x; }

 7  * }

 8  */

 9 public class Solution {

10     public ListNode InsertionSortList(ListNode head) {

11         if (head == null) return null;

12         ListNode hpre = new ListNode(0);

13         hpre.next = head;

14         ListNode pre = head;

15         head = head.next;

16         while (head != null) {

17             if (head.val > pre.val) {

18                 head = head.next;

19                 pre = pre.next;

20             }

21             else {

22                 pre.next = head.next;

23                 ListNode tmp = hpre;

24                 while (head.val > tmp.next.val) tmp = tmp.next;

25                 head.next = tmp.next;

26                 tmp.next = head;

27                 head = pre.next;

28             }

29         }

30         return hpre.next;

31     }

32 }

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