刷題過程當中經常遇到排序問題,Java中自帶的sort方法能夠很是方便的幫助咱們進行排序。html
常見的排序問題有兩種情形:java
1.對一個數組進行排序。git
2.對自定義類型的類進行排序。數組
一般狀況下咱們可使用Array.sort()來對數組進行排序,有如下3種狀況:this
1.Array.sort(int[] a)spa
直接對數組進行升序排序code
2.Array.sort(int[] a , int fromIndex, int toIndex)orm
對數組的從fromIndex到toIndex進行升序排序htm
3.新建一個comparator從而實現自定義比較blog
具體方法以下:
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"); //要實現減序排序,得經過包裝類型數組,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]+" "); } } }
當咱們處理自定義類型的排序時,通常將自定義類放在List種,以後再進行排序
通常咱們對自定義類型數據進行重寫Comparator來進行對數據進行比較
具體方法以下:
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>(){ //咱們但願對自定義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:
Excel can sort records according to any column. Now you are supposed to imitate this function.
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).
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.
3 1 000007 James 85 000010 Amy 90 000001 Zoe 60
000001 Zoe 60 000007 James 85 000010 Amy 90
4 2 000007 James 85 000010 Amy 90 000001 Zoe 60 000002 James 98
000010 Amy 90 000002 James 98 000007 James 85 000001 Zoe 60
4 3 000007 James 85 000010 Amy 90 000001 Zoe 60 000002 James 90
000001 Zoe 60 000007 James 85 000002 James 90 000010 Amy 90
顯然咱們新建的類中應該包含ID,name,val(成績) ,以後咱們新建三種比較器即可以完成這三種比較方式了,代碼以下:
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:
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.
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.
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.
2 5 1234567890001 95 1234567890005 100 1234567890003 95 1234567890002 77 1234567890004 85 4 1234567890013 65 1234567890011 25 1234567890014 100 1234567890012 85
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
這到題目須要加入兩種排序方式——考試地點的排名,以及總排名,因此這裏我在每個考場都創建了一個list進行排序,還有一個總的ranking對全部考生進行排序。
代碼以下:
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 ; } } }