2018年10月31日

隊(duì)列是一種先進(jìn)先出(FIFO)的數(shù)據(jù)結(jié)構(gòu)

1够滑，隊(duì)列的鏈表實(shí)現(xiàn)

public class ListQueue<Item> implements Iterable<Item> {
    private class Node {
        Item item;
        Node next;
    }

    private Node first;
    private Node last;
    private int N;

    public boolean isEmpty() {
        return first == null;
    }

    public int size() {
        return N;
    }

    public void enqueue(Item item) {
        Node oldLast = last;
        last = new Node();
        last.item = item;
        last.next = null;
        if (isEmpty()) {
            first = last;
        } else {
            oldLast.next = last;
        }
        N++;
    }

    public Item dequeue() {
        Item item = first.item;
        first = first.next;
        if (isEmpty()) {
            last = null;
        }
        N--;
        return item;
    }

    @Override
    public Iterator<Item> iterator() {
        return new ListIterator(first);
    }

    private class ListIterator implements Iterator<Item> {
        private Node current;

        public ListIterator(Node first) {
            current = first;
        }

        @Override
        public boolean hasNext() {
            return current != null;
        }

        @Override
        public void remove() {
            throw new UnsupportedOperationException();
        }

        @Override
        public Item next() {
            if (!hasNext()) {
                throw new NoSuchElementException();
            }
            Item item = current.item;
            current = current.next;
            return item;
        }
    }
}

2添怔，隊(duì)列的數(shù)組實(shí)現(xiàn)

public class ResizingArrayQueue<Item> implements Iterable<Item> {
    private Item[] a = (Item[]) new Object[2];
    private int N;
    private int first;
    private int last;

    public boolean isEmpty() {
        return N == 0;
    }

    public int size() {
        return N;
    }

    private void resize(int max) {
        Item[] temp = (Item[]) new Object[max];
        for (int i = 0; i < N; i++) {
            temp[i] = a[(first + i) % a.length];
        }
        a = temp;
        first = 0;
        last = N;
    }

    public void enqueue(Item item) {
        if (N == a.length) {
            resize(2 * a.length);
        }
        a[last++] = item;
        if (last == a.length) {
            //環(huán)形數(shù)組,到底了從頭計(jì)數(shù)
            last = 0;
        }
        N++;
    }

    public Item dequeue() {
        if (isEmpty()) {
            throw new NoSuchElementException();
        }
        Item item = a[first];
        //避免對(duì)象游離续徽，即保存一個(gè)不需要的對(duì)象的引用
        a[first] = null;
        first++;
        N--;
        if (first == a.length) {
            //環(huán)形數(shù)組畏妖，到底了從頭開始
            first = 0;
        }
        if (N > 0 && N == a.length / 4) {
            resize(a.length / 2);
        }
        return item;
    }

    @Override
    public Iterator<Item> iterator() {
        return new ArrayIterator();
    }

    private class ArrayIterator implements Iterator<Item> {
        private int i = 0;

        @Override
        public boolean hasNext() {
            return i < N;
        }

        @Override
        public void remove() {
            throw new UnsupportedOperationException();
        }

        @Override
        public Item next() {
            if (!hasNext()) {
                throw new NoSuchElementException();
            }
            Item item = a[(first + i) % a.length];
            i++;
            return item;
        }
    }
}

3，隊(duì)列的應(yīng)用

• 圓圈中最后剩下的數(shù)字
  題目：0, 1, …, n-1這n個(gè)數(shù)字排成一個(gè)圓圈踢代，從數(shù)字0開始每次從這個(gè)圓圈里刪除第m個(gè)數(shù)字盲憎。求出這個(gè)圓圈里剩下的最后一個(gè)數(shù)字。

   public int lastRemainingSolution(int n, int m) {
       Queue<Integer> queue = new LinkedList<>();
       for (int i = 0; i < n; i++) {
           queue.offer(i);
       }
       int count = 0;
       int result = -1;
       while (!queue.isEmpty()) {
           if (count == m - 1) {
               result = queue.poll();
               count = 0;
           }
           if (!queue.isEmpty()) {
               queue.offer(queue.poll());
           }
           count++;
       }
       return result;
   }