题目链接
分析
我们需要排序,不如用TreeMap或者TreeSet,但有三个关键词,所以确实不好办。
斟酌再三,决定用TreeSet,而且将三个属性完全封装为一个类而不是一个1对2(元素数量)的映射关系,反而简单一些。
这样,我们利用面向对象的类,简化了问题的复杂度:
private static class Grade {
Integer num;
Integer chinese;
Integer grade;
public Grade(int num, int chinese, int grade) {
this.num = num;
this.chinese = chinese;
this.grade = grade;
}
}
我们根本不关注数学和英语成绩(在总分计算之后),所以不需要存储。
最后输出的话,因为题意暗示不会出现小于5人的状况,所以不需要担心,加一个counter,到5就break循环就好啦。
TreeMap需要自己制定排序规则,如下:
Set<Grade> set = new TreeSet<>(new Comparator<Grade>() {
@Override
public int compare(Grade grade1, Grade grade2) {
int result1 = -grade1.grade.compareTo(grade2.grade);
if (result1 != 0) {
return result1;
}
int result2 = -grade1.chinese.compareTo(grade2.chinese);
if (result2 != 0) {
return result2;
}
return grade1.num.compareTo(grade2.num);
}
});
正负号一定要弄明白哈,别搞错啦!!
AC代码(Java语言描述)
import java.util.*;
public class Main {
private static class Grade {
Integer num;
Integer chinese;
Integer grade;
public Grade(int num, int chinese, int grade) {
this.num = num;
this.chinese = chinese;
this.grade = grade;
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int num = Integer.parseInt(scanner.nextLine());
Set<Grade> set = new TreeSet<>(new Comparator<Grade>() {
@Override
public int compare(Grade grade1, Grade grade2) {
int result1 = -grade1.grade.compareTo(grade2.grade);
if (result1 != 0) {
return result1;
}
int result2 = -grade1.chinese.compareTo(grade2.chinese);
if (result2 != 0) {
return result2;
}
return grade1.num.compareTo(grade2.num);
}
});
for (int i = 1; i <= num; i++) {
String[] array = scanner.nextLine().split(" ");
int chinese = Integer.parseInt(array[0]);
int grade = chinese + Integer.parseInt(array[1]) + Integer.parseInt(array[2]);
set.add(new Grade(i, chinese, grade));
}
int counter = 0;
for (Grade grade : set) {
counter++;
System.out.println(grade.num + " " + grade.grade);
if (counter == 5) {
break;
}
}
scanner.close();
}
}