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mit java open course assignment #2

2015-09-12 12:28 555 查看
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);
}
}


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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
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