Java的。 如何将两个最初排序的队列合并到一个队列中？ （没有链接列表或其他内容）

Question

For example I have first queue: front[1 3 5 7]back and second queue front[2 4 6 8]back. New merged queue must be: front[1 2 3 4 5 6 7 8]back. Or first queue: front[1 3 3 7]back, second: front[2 4 5 7 8]back. New merged queue: front[1 2 3 3 4 5 7 7 8]back. And we can't use any types of sorting. Here is the my implementation of queue:

public class QueueImpl implements IntQueue {

    protected int capacity;
    public static final int CAPACITY = 100;
    protected int Q[];
    protected int f = 0, r = 0;

    public QueueImpl() {
        this(CAPACITY);
    }

    public QueueImpl(int cap) {
        capacity = cap;
        Q = new int[capacity];
    }


    public void enqueue(int value) throws Exception {
        if (getSize() == capacity - 1) {
            throw new Exception("Full"); //To change body of generated methods, choose Tools | Templates.
        }

        Q[r] = value;

        r = (r + 1) % capacity;

    }


    public int dequeue() throws Exception {
        int element;
        if (isEmpty()) {
            throw new Exception("Empty"); //To change body of generated methods, choose Tools | Templates.
        }
        element = Q[f];

        f = (f + 1) % capacity;
        return element;

    }


    public int getSize() {
        return (capacity - f + r) % capacity;
    }


    public boolean isEmpty() {
        return (f == r);
    }

    public String toString() {
        int z = f;
        String s;
        s = "f[";

        if (getSize() >= 1) {
            s += Q[0];
            for (int i = 1; i <= getSize() - 1; i++) {
                s += " " + Q[z + 1];
                z = (z + 1) % capacity;

            }
        }
        return s + "]b";
    }

}

My solution: public class Assign2Problem3Solution {

public static IntQueue merge(IntQueue q1, IntQueue q2) throws Exception {
    IntQueue merged = new QueueImpl();
    int a, b, k, t, m;




    if (a < b) {
        k = a;
        t = b - a;
    } else {
        k = b;
        t = a - b;
    }

    for (int i = 0; i < k; i++) {

        a = q1.dequeue();
        b = q2.dequeue();
        if (a < b) {
            merged.enqueue(a);
            merged.enqueue(b);
        } else if (b < a) {
            merged.enqueue(b);
            merged.enqueue(a);
        }

    }
    if (q1.getSize() > q2.getSize()) {

        for (int i = 0; i < t; i++) {
            m = q1.dequeue();
            merged.enqueue(m);

        }
    } else if (q1.getSize() < q2.getSize()) {
        for (int i = 0; i < t; i++) {
            m = q2.dequeue();
            merged.enqueue(m);

        }
    }

    return merged;
}

}

Here is the code that works and satisfies the conditions:

IntQueue merged = new QueueImpl(); int a, b;

    if (!q1.isEmpty() && !q2.isEmpty()) {
        a = q1.dequeue();
        b = q2.dequeue();
        while (true) {
            if (a < b) {
                merged.enqueue(a);
                if (q1.isEmpty()) {
                    merged.enqueue(b);
                    break;
                }
                a = q1.dequeue();
            } else {
                merged.enqueue(b);
                if (q2.isEmpty()) {
                    merged.enqueue(a);
                    break;
                }
                b = q2.dequeue();
            }
        }
    }
    if (q1.getSize() > 0) {
        while (!q1.isEmpty()) {
            a = q1.dequeue();
            merged.enqueue(a);
        }
    } else if (q2.getSize() > 0) {
        while (!q2.isEmpty()) {
            b = q2.dequeue();
            merged.enqueue(b);
        }
    } 

    return merged;

Answer 1

The problem is here:

    a = q1.dequeue();
    b = q2.dequeue();
    if (a < b) {
        merged.enqueue(a);
        merged.enqueue(b);
    } else if (b < a) {
        merged.enqueue(b);
        merged.enqueue(a);
    }

This code means remove one element from the first queue, and remove one element from the second queue. Add the smaller element into the merged queue, and then add the larger element into the merged queue.

The above code does not work for some cases. One example is this. Consider two queues Q1 = {1, 2, 3} and Q2 = {4, 5, 6}. In step 1 (loop, k = 0), we remove 1 from Q1 and 4 from Q2. Because 1 is smaller than 4, we add 1, followed by 4. The merged queue is {1, 4}, Q1 is now {2, 3}, and Q2 is now {5, 6}. In step 2 (loop, k = 1), we remove 2 from Q1 and 5 from Q2. Because 2 is smaller than 5, we add 2, followed by 5. The merged queue is now {1, 4, 2, 5}. Notice that although 2 is smaller than 4, we add 2 after 4, which is incorrect. The problem here is that in step 1, we cannot immediately add 4, because the next element in Q1 (which is 2) may be smaller than 4.

What you need to do is something like this:

int e1 = q1.dequeue();
int e2 = q2.dequeue();

while (true) {
  if (e1 < e2) {
    merged.enqueue(e1);
    if (q1.isEmpty()) {
      // add remaining q2 elements
      while (!q2.isEmpty()) { 
        merged.enqueue(q2.dequeue());
      }
      break;
    }
    // take another element from q1
    e1 = q1.dequeue();
  } else {
    merged.enqueue(e2);
    if (q2.isEmpty()) {
      // add remaining q1 elements
      while (!q1.isEmpty()) { 
        merged.enqueue(q1.dequeue());
      }
      break;
    }
    // take another element from q2
    e2 = q2.dequeue();
  }
}

If you have a method that can retrieve the head element, without removing it from the queue, the code can be much cleaner.

Answer 2

Implement the peek() method, and compare the values as they are peeked to see which on is smaller and start adding them into a new queue.

Pretty much the merge part of a "MergeSort"

Answer 3

You could simply do the following :

C = new queue
a = A.poll() 
b = B.poll()
While A or B is NOT EMPTY
    while( b < a ) 
        - C.put(b)
        - b = B.Poll()
    - C.put(a)
    - a = A.poll()
FILL C with the queue left 

It is just alterning between both list taking each time the number closest to the last number in C.At the end, you just fill the last number from the queue that is not empty.

You got to pay attention to empty list, here it s just a really simple pseudo code, but when i do "b = B.Poll()" i should pay attention whether there is still an element in B or not, otherwise it might throw an exception.

Well, i've just tried this on a simple example, i do not say it is working 100% but it gives you ideas. Ps: here my "poll" => "dequeue" for your class