mit java open course assignment #2

package come;

public class Marothon {

    public static void FirstName(String[] args1,int[] args2){
        int MinIndex,MinTime,i;
        MinTime = 10000;
        MinIndex = 0;
        for(i = 0; i < 16; i++){
            if (args2[i] < MinTime){
                MinTime = args2[i];
                MinIndex = i;
            }
        }
        System.out.println(args1[MinIndex] + " "+MinTime);
    }
    public static void SecondName(String[] args1, int[] args2){
        int SmallIndex,SmallerIndex,Small,Smaller,i;
        Small = 10000;
        Smaller = 10000;
        SmallerIndex = 100;
        SmallIndex =101;
        for(i = 0; i<16; i++){
            if(args2[i] < Small && args2[i] > Smaller){
                Small = args2[i];
                SmallIndex = i;
            }else if(args2[i] < Smaller){
                Small = Smaller;
                Smaller = args2[i];
                SmallIndex = SmallerIndex;
                SmallerIndex = i;
            }
        }
        System.out.println(args1[SmallIndex] + " " + args2[SmallIndex]);
    }
    public static void main(String[] args){
        // TODO Auto-generated method stub
        String[] names = {"Elena", "Thomas", "Hamilton", "Suzie", "Phil", "Matt", "Alex",
                "Emma", "John", "James", "Jane", "Emily", "Daniel", "Neda",
                "Aaron", "Kate"};
        int[] times = {341, 273, 278, 329, 445, 402, 388, 275, 243, 334, 412, 393, 299,
                343, 317, 265};
        SecondName(names,times);
}
}

SecondName函数只遍历了一次数组，大大提升了之前想用FirstName遍历一次后用if避开最小的来做，不过就要遍历两次，哈哈哈，不错