如何在java中使用二分搜索找到排名2的学生？

Question

In an exam that was attended by 400 students, the teacher wanted to give a special gift to the student who was ranked 2. If all participants are identified by a test number (1-400) create the MOST EFFICIENT algorithm and program to find students who will be rewarded!

*) The same value has the same rank (for example 100 100 98 98 then there are 2 people in the 1st rank and 2 people in the 2nd rank)

note: Help me, how is the code in java, i really confuse :(

Answer 1

You don't need to sort the input, simply:

1. store current top rank, and the list of people in it (initially null/empty)
2. store current second top rank, and the list of people in it (initially null/empty)

loop through the items and:

1. if current rank > top rank => second top rank is replaced by current top rank, and current top rank is replaced by (rank/the current item)
2. if current rank = top rank => push student onto current top rank list
3. if current rank < top rank => do checks 1 and 2, but this time for second rank

For instance, assume the items are Joe:1 Jack:2 Jill:3, Jim: 1, Job: 2 then:

Joe ranks higher than null => current top rank is 1 and items are [Joe]

Jack ranks higher than current top (1) => set top rank to 2 and items to [Jack], and also set second top to previous top, ie 1 and [Joe]

Jill ranks higher than current top (2) => set top rank to 3 and items to [Jill], and also set second top to previous top, ie 2 and [Jack]

Jim ranks lower than current top (3) and lower than second top (2) => nothing to do

Job ranks lower than current top (3) but is equal to second top (2) => append Job to second top list, so it is [Jack and Job]

The result is top: 3 and [Jill] and second top: 2 and [Jack and Job].

If the input is sorted, you can use a binary search, but sorting is more expensive (n log n) than just doing the above.

Answer 2

You must have two arrays : one for the identifying numbers of students and the other with their score Lets name : "a" the score array and "b" the identifying numbers array

This will have you arrays sorted:

public static void bubbleSort(int[] a,int[]b) {
        boolean sorted = false;
        int temp;
        while(!sorted) {
            sorted = true;
            for (int i = 0; i < a.length - 1; i++) {
                if (a[i] > a[i+1]) {
                    temp = a[i];
                    a[i] = a[i+1];
                    a[i+1] = temp;
                    temp = b[i];
                    b[i] = b[i+1];
                    b[i+1] = temp;
                    sorted = false;
                }
            }
        }
    }

after you short them find the second biggest number with this

   static int second(int []a){
        int number =a[0];//case all of them had same score
        for(int i=0;i<a.length-1;i++){
            if (a[i]>a[i+1]){
                number= a[i+1];
                break;
            }
        }
        return number;
    }

and now store the result in a variable number and:

for(int i=0;i<a.length;i++){
            if (a[i]==number){
                System.out.println(b[i]);
            }
        }

Of course there are more solutions but this is mine...