Java基础算法--排序

一、插入排序

　　数组：　　

 1 void insertionSort(int[] a){

 2     int j;

 3     for(int p = 1;p<a.length;p++){

 4         int tmp = a[p];

 5         for(j=p;j>0;j--){

 6             if(tmp<a[j-1]){

 7                 a[j] = a[j-1];

 8             }else{

 9                 a[i] = tmp;

10             }

11         }    

12     }

13 }

　　若是Java泛型掌握的好，可以这样书写

1 void insertionSort(AnyType[] a){

2     int j;

3     for(int p = 1;p < a.length;p++){

4         AnyType tmp = a[p];

5         for(j=p;j>0&&tmp.compareTo(a[j-1])<0;j--)            

6             a[j] = a[j-1];    

7         a[i] = tmp;    

8     }    

9 }

　　链表：

 1 public class Solution {

 2     public ListNode insertionSortList(ListNode head) {

 3         ListNode dummy = new ListNode(-1);

 4         dummy.next = head;

 5         head = dummy;

 6         ListNode p = dummy.next;

 7         ListNode s;

 8         

 9         while(p!=null){

10             if(p.next==null)  break;

11             while(head!=p&&p.next.val>head.next.val){

12                 head=head.next;

13             }

14             if(head==p)

15                 p=p.next;

16             else{

17                 s = p.next;

18                 p.next = p.next.next;

19                 s.next = null;

20                 

21                 ListNode tmp = head.next;

22                 head.next = s;

23                 s.next = tmp;

24             }

25             head=dummy;

26         }

27         return dummy.next;

28     }

29 }

 另一种于九章算法上的做法：

 1 public class Solution {

 2     public ListNode insertionSortList(ListNode head) {

 3         ListNode dummy = new ListNode(0);

 4 

 5         while (head != null) {

 6             ListNode node = dummy;

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

 8                 node = node.next;

 9             }

10             ListNode temp = head.next;

11             head.next = node.next;

12             node.next = head;

13             head = temp;

14         }

15 

16         return dummy.next;

17     }

18 }