以奇数索引反转队列中的数字

Question

This is aa review question(not asking for HW answers!!!)

Write a method reverseHalf that reverses the order of half of the elements of a Queue of integers. Your method should reverse the order of all the elements in odd-numbered positions (position 1, 3, 5, etc.) assuming that the first value in the queue has position 0. For example, if the queue originally stores this sequence of numbers when the method is called:

index: 0 1 2 3 4 5 6 7 front [1, 8, 7, 2, 9, 18, 12, 0] back - it should store the following values after the method finishes executing:

index: 0 1 2 3 4 5 6 7 front [1, 0, 7, 18, 9, 2, 12, 8] back Notice that numbers in even positions (positions 0, 2, 4, 6) have not moved. That sub-sequence of numbers is still: (1, 7, 9, 12). But notice that the numbers in odd positions (positions 1, 3, 5, 7) are now in reverse order relative to the original. In other words, the original sub-sequence: (8, 2, 18, 0) - has become: (0, 18, 2, 8). You may use a single stack as auxiliary storage.

I have gotten it to work with an even number of elements in my queue, but am lost when it comes to an odd number....do I have to re-structure my entire program? Or is this one of those quick-fixes??

My code:

public static void reverseHalf(Queue<Integer> q){
    int size = q.size();
    Stack<Integer> s = new Stack<Integer>();


    for(int i = 0; i < size; i++){
        int n = q.remove();
        s.push(n);
        q.add(n);
    }
    for(int i = 0; i < size; i++){
        int n = q.remove();
        if(i % 2 == 0){
            q.add(n);
        }
    }

    for(int i = 0; i < size; i++){
        int n = s.pop();
        if(i % 2 == 0)
        q.add(n); 
    }

    for(int i = 0; i < size; i++){
        int n = q.remove();
        s.push(n);
        q.add(n);
    }

    for(int i = (size / 2); i < size; i++){
        q.remove();
    }

    for(int i = 0; i < (size / 2); i++){
        s.pop();
    }

    for(int i = 0; i < size / 2; i++){
        q.add(s.pop());
    }

    s.clear();

    for(int i = 0; i < size; i++){
        int n = q.remove();
        s.push(n);
        q.add(n);
    }

    for(int i = 0; i < (size  / 2); i++){
        int n = s.pop();
        int m = q.remove();
        q.add(n);
        q.add(m);
    }

    for(int i = 0; i < (size / 2); i++){
        q.remove();
    }


}

Answer 1

First of all you need to replace all of the

i % 2 == 0

with

i % 2 != 0

The above is the definition of an odd number. The second trick lies in all of the lines containing

for(int i = 0; i < (size  / 2); i++)

you should be able to figure this one out easily.

Answer 2

Initialize i to 1 and j to the last odd index. Swap the elements at these index. Increment i by 2 and decrement j by 2 until j Less than i. I think this should would work.