刷題過程中常常遇到排序問題，Java中自帶的sort方法可以非常友善的幫助我們進行排序。
常見的排序問題有兩種情形：
1.對一個數組進行排序。
2.對自定義類型的類進行排序。
一，對數組進行排序：
通常情況下我們可以使用Array.sort()來對數組進行排序，有以下3種情況：
1.Array.sort(int[] a)
直接對數組進行升序排序
2.Array.sort(int[] a , int fromIndex, int toIndex)
對數組的從fromIndex到toIndex進行升序排序
3.建立一個comparator進而實作自定義比較
具體方法如下：
import java.util.*;
public class no {
public static void main(String []args)
{
int[] ints=new int[]{2,324,4,57,1};
System.out.println("增序排序後順序");
Arrays.sort(ints);
for (int i=0;i<ints.length;i++)
{
System.out.print(ints[i]+" ");
}
System.out.println("\n減序排序後順序");
//要實作減序排序，得通過包裝類型數組，基本類型數組是不行滴
Integer[] integers=new Integer[]{2,324,4,4,6,1};
Arrays.sort(integers, new Comparator<Integer>()
{
public int compare(Integer o1, Integer o2)
{
return o2-o1;
}
public boolean equals(Object obj)
{
return false;
}
});
for (Integer integer:integers)
{
System.out.print(integer+" ");
}
System.out.println("\n對部分排序後順序");
int[] ints2=new int[]{212,43,2,324,4,4,57,1};
//對數組的[2,6)位進行排序
Arrays.sort(ints2,2,6);
for (int i=0;i<ints2.length;i++)
{
System.out.print(ints2[i]+" ");
}
}
} 二，對自定義類進行排序
當我們處理自定義類型的排序時，一般将自定義類放在List種，之後再進行排序
一般我們對自定義類型資料進行重寫Comparator來進行對資料進行比較
具體方法如下：
public static class Adam
{
int ID ;
int val ;
String name ;
Adam(int ID , String name , int val)
{
this.ID = ID ;
this.name = name ;
this.val = val ;
}
}
Collections.sort(list, new Comparator<Object>(){ //我們希望對自定義Adam中的ID進行排序
public int compare(Object a , Object b)
{
Adam student1 = (Adam)a ;
Adam student2 = (Adam)b ;
return student1.ID - student2.ID ;
}
}); 下面來分析兩道題目：
1028 List Sorting （25 分）
Excel can sort records according to any column. Now you are supposed to imitate this function.
Input Specification:
Each input file contains one test case. For each case, the first line contains two integers N (≤105) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student's record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).
Output Specification:
For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID's; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID's in increasing order.
Sample Input 1:
3 1
000007 James 85
000010 Amy 90
000001 Zoe 60
Sample Output 1:
000001 Zoe 60
000007 James 85
000010 Amy 90
Sample Input 2:
4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98
Sample Output 2:
000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60
Sample Input 3:
4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90
Sample Output 3:
000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90
顯然我們建立的類中應該包含ID，name，val(成績) ，之後我們建立三種比較器便可以完成這三種比較方式了，代碼如下：
import java.util.*;
public class ListSorting2 {
public static void main(String args[])
{
Scanner scanner = new Scanner(System.in) ;
int nums = scanner.nextInt() ;
int choose = scanner.nextInt() ;
scanner.nextLine() ;
List<Adam> list = new ArrayList<>() ;
for(int i = 0 ; i < nums ; i ++)
{
String s = scanner.nextLine() ;
String[] s1 = s.split(" ") ;
Adam student = new Adam(Integer.parseInt(s1[0]),s1[1],Integer.parseInt(s1[2])) ;
list.add(student) ;
}
scanner.close();
if(choose == 1)
Collections.sort(list, new Comparator<Object>(){
public int compare(Object a , Object b)
{
Adam student1 = (Adam)a ;
Adam student2 = (Adam)b ;
return student1.ID - student2.ID ;
}
});
else if(choose == 2)
Collections.sort(list,new Comparator<Object>(){
public int compare(Object a , Object b)
{
Adam student1 = (Adam)a ;
Adam student2 = (Adam)b ;
int n = student1.name.compareTo(student2.name) ;
if(n == 0)
{
return student1.ID - student2.ID ;
}
else return n ;
}
});
else
{
Collections.sort(list,new Comparator<Object>(){
public int compare(Object a , Object b)
{
Adam student1 = (Adam)a ;
Adam student2 = (Adam)b ;
int n = student1.val - student2.val ;
if(n == 0) return student1.ID - student2.ID ;
else
return n ;
}
});
}
for(int i = 0 ; i < list.size() ; i ++)
{
System.out.printf("%06d",list.get(i).ID);
System.out.println(" "+list.get(i).name+" "+list.get(i).val);
}
}
public static class Adam
{
int ID ;
int val ;
String name ;
Adam(int ID , String name , int val)
{
this.ID = ID ;
this.name = name ;
this.val = val ;
}
}
} 再看一道題目：
1025 PAT Ranking （25 分）
Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive number N (≤100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (≤300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.
Output Specification:
For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:
registration_number final_rank location_number local_rank
The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.
Sample Input:
2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85
Sample Output:
9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4
這到題目需要加入兩種排序方式——考試地點的排名，以及總排名，是以這裡我在每一個考場都建立了一個list進行排序，還有一個總的ranking對所有考生進行排序。
代碼如下：
import java.util.*;
public class PATRanking {
public static void main(String args[])
{
Scanner scanner = new Scanner(System.in) ;
int locals = scanner.nextInt() ;
int sum = 0 ;
@SuppressWarnings("unchecked")
List<Adam>[] list = new List[locals] ;
List<Adam> ranking = new ArrayList<>() ;
for(int i = 0 ; i < locals ; i ++)
{
list[i] = new ArrayList<>() ;
int nums = scanner.nextInt() ;
sum += nums ;
scanner.nextLine() ;
for(int j = 0 ; j < nums ; j ++)
{
String s = scanner.nextLine() ;
String[] s1 = s.split(" ") ;
Adam a = new Adam(s1[0],Integer.parseInt(s1[1]),i+1) ;
list[i].add(a) ;
ranking.add(a) ;
}
}
scanner.close();
for(int i = 0 ; i < locals ; i ++)
{
Collections.sort(list[i], new Comparator<Object>(){
public int compare(Object a , Object b)
{
Adam student1 = (Adam)a ;
Adam student2 = (Adam)b ;
int n = student2.val - student1.val ;
if(n == 0) return student1.ID.compareTo(student2.ID) ;
else return n ;
}
});
for(int j = 0 ; j < list[i].size() ; j ++)
{
if(j == 0)
{
list[i].get(j).localrank = j + 1 ;
continue ;
}
if(list[i].get(j).val == list[i].get(j-1).val)
list[i].get(j).localrank = list[i].get(j-1).localrank ;
else
list[i].get(j).localrank = j + 1 ;
}
}
Collections.sort(ranking,new Comparator<Object>(){
public int compare(Object a , Object b)
{
Adam student1 = (Adam)a ;
Adam student2 = (Adam)b ;
int n = student2.val - student1.val ;
if(n == 0) return student1.ID.compareTo(student2.ID) ;
else return n ;
}
});
System.out.println(sum);
for(int i = 0 ; i < ranking.size() ; i ++)
{
Adam a = ranking.get(i) ;
if(i == 0)
{
a.rank = i + 1 ;
System.out.println(a.ID+" "+a.rank+" "+a.local+" "+a.localrank);
continue ;
}
if(ranking.get(i).val == ranking.get(i-1).val)
ranking.get(i).rank = ranking.get(i-1).rank ;
else
a.rank = i + 1 ;
System.out.println(a.ID+" "+a.rank+" "+a.local+" "+a.localrank);
}
}
public static class Adam
{
String ID ;
int val ;
int rank ;
int localrank ;
int local ;
Adam(String ID , int val , int local)
{
this.ID = ID ;
this.val = val ;
this.local = local ;
}
}
} ![Java排序算法---快速排序[圖]](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACwAAAAAAQABAAACAkQBADs=)