以下是我的我只用main函數(shù)編的代碼:
package?com.jieduan1;
import?java.util.Arrays;
public?class?demo06?{
????
????//完成?main?方法
????public?static?void?main(String[]?args)?{
????????int?[]?scores1={89,-23,64,91,119,52,73};
????????int?n=0;
????????for?(int?i=0;i<scores1.length;i++)?{
???????? if?(scores1[i]>=0&&scores1[i]<=100)?{
???????? n++;
???????? }
????????}
????????System.out.println("有效成績個數(shù)為:"+n);
????????int?[]?scores2=new?int[n];
????????int?j=0;
????????for?(int?i=0;i<scores1.length;i++)?{
???????? if?(scores1[i]>=0&&scores1[i]<=100)?{
???????? scores2[j]=scores1[i];
???????? j++;
???????? }
????????
????????}
????????System.out.println("輸出有效成績:"+Arrays.toString(scores2));
????????Arrays.sort(scores2);
????????System.out.println("輸出排序后的有效成績:"+Arrays.toString(scores2));
????????int?[]?scores3=new?int?[3];
????????int?k=3-1;
????????for?(int?i=n-3;i<n;i++)?{
???????? scores3[k]=scores2[i];
???????? k--;
????????}
????????System.out.println("輸出前三名成績:"+Arrays.toString(scores3));
????????for?(int?i=0;i<3;i++)?{
???????? System.out.println("考試成績的第"+(i+1)+"名為:"+scores3[i]);
????????}
????????
????????
????????
????}
????
????
????
????
????
????
????
}
執(zhí)行結(jié)果為:
有效成績個數(shù)為:5
輸出有效成績:[89,?64,?91,?52,?73]
輸出排序后的有效成績:[52,?64,?73,?89,?91]
輸出前三名成績:[91,?89,?73]
考試成績的第1名為:91
考試成績的第2名為:89
考試成績的第3名為:73
2017-03-23
自定義方法可以拆分邏輯,實現(xiàn)代碼的復(fù)用,你貼出來的這段代碼只是一個簡單練習(xí),也就無所謂拆分復(fù)用了。
如果不理解,可以暫時不去糾結(jié)這個問題,以后敲的代碼多了你就有體會了