3、动态数组

内容来自刘宇波老师算法与数据结构体系课

1、动态数组

在这里新创建一个数组类,对 Java 语言中的原始数组进行封装,使得它可以动态的扩容和缩容
Java 语言中也有类似的实现,叫 ArrayList,我们创建的数组类是它的简化版本,下面是代码实现
/**
 * 动态数组
 */
public class Array<E> {

    private E[] data;
    private int size;

    public Array(int capacity) {
        data = (E[]) new Object[capacity];
        size = 0;
    }

    public Array() {
        this(10);
    }

    public Array(E[] arr) {
        data = Arrays.copyOf(arr, arr.length);
        size = arr.length;
    }

    public int getSize() {
        return size;
    }

    public int getCapacity() {
        return data.length;
    }

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

    /**
     * 添加
     */
    public void add(int index, E e) {
        if (index < 0 || index > size) throw new RuntimeException("need 0 <= index <= size");
        if (size == data.length) resize(data.length * 2);

        System.arraycopy(data, index, data, index + 1, size - index);
        data[index] = e;
        size++;
    }

    public void addFirst(E e) {
        add(0, e);
    }

    public void addLast(E e) {
        add(size, e);
    }

    /**
     * 删除
     */
    public E remove(int index) {
        if (index < 0 || index >= size) throw new RuntimeException("need 0 <= index < size");

        E ret = data[index];
        System.arraycopy(data, index + 1, data, index, size - index - 1);
        size--;
        data[size] = null;

        if (size == data.length / 4 && data.length / 2 != 0) resize(data.length / 2);
        return ret;
    }

    public E removeFirst() {
        return remove(0);
    }

    public E removeLast() {
        return remove(size - 1);
    }

    public void removeElement(E e) {
        int index = find(e);
        if (index != -1) remove(index);
    }

    /**
     * 修改
     */
    public void set(int index, E e) {
        if (index < 0 || index > size) throw new RuntimeException("need 0 <= index <= size");
        data[index] = e;
    }

    /**
     * 查看
     */
    public E get(int index) {
        if (index < 0 || index >= size) throw new RuntimeException("need 0 <= index < size");
        return data[index];
    }

    public E getFirst() {
        return get(0);
    }

    public E getLast() {
        return get(size - 1);
    }

    public boolean contains(E e) {
        for (int i = 0; i < size; i++) {
            if (data[i].equals(e)) return true;
        }
        return false;
    }

    public int find(E e) {
        for (int i = 0; i < size; i++) {
            if (data[i].equals(e)) return i;
        }
        return -1;
    }

    /**
     * 动态数组
     */
    private void resize(int newCapacity) {
        E[] newData = (E[]) new Object[newCapacity];
        System.arraycopy(data, 0, newData, 0, size);
        data = newData;
    }

    public void swap(int a, int b) {
        if (a < 0 || a >= size || b < 0 || b >= size) {
            throw new IllegalArgumentException("Swap failed, require 0 <= index < size");
        }
        E k = data[a];
        data[a] = data[b];
        data[b] = k;
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append(String.format("Array: size = %d, capacity = %d\n", size, data.length));
        sb.append('[');
        for (int i = 0; i < size; i++) {
            sb.append(data[i]);
            if (i != size - 1) sb.append(", ");
        }
        sb.append(']');
        return sb.toString();
    }
}

2、resize 复杂度分析

image

3、复杂度震荡

image

4、数组栈

很容易基于上面的动态数组来实现栈,并且 push、pop、peek 的复杂度都是 O(1) 级别的
/**
 * 数组栈
 */
public interface Stack<E> {

    void push(E e);

    E pop();

    E peek();

    int getSize();

    boolean isEmpty();

}
public class ArrayStack<E> implements Stack<E> {

    private final Array<E> array;

    public ArrayStack() {
        this.array = new Array<>();
    }

    public ArrayStack(int capacity) {
        this.array = new Array<>(capacity);
    }

    @Override
    public void push(E e) {
        array.addLast(e);
    }

    @Override
    public E pop() {
        return array.removeLast();
    }

    @Override
    public E peek() {
        return array.getLast();
    }

    @Override
    public int getSize() {
        return array.getSize();
    }

    @Override
    public boolean isEmpty() {
        return array.isEmpty();
    }

    public int getCapacity() {
        return array.getCapacity();
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append("ArrayStack: [");
        for (int i = 0; i < array.getSize(); i++) {
            sb.append(array.get(i));
            if (i != array.getSize() - 1) sb.append(", ");
        }
        sb.append("] Top");
        return sb.toString();
    }
}

5、数组队列

也很容易基于上面的动态数组来实现队列,enqueue 和 getFront 的复杂度都是 O(1) 级别的,但 dequeue 却是 O(n)
或许可以优化,使得 dequeue 的复杂度也是 O(1) 级别的
/**
 * 数组队列
 */
public interface Queue<E> {

    void enqueue(E e);

    E dequeue();

    E getFront();

    int getSize();

    boolean isEmpty();

}
public class ArrayQueue<E> implements Queue<E> {

    private final Array<E> array;

    public ArrayQueue() {
        this.array = new Array<>();
    }

    public ArrayQueue(int capacity) {
        this.array = new Array<>(capacity);
    }

    @Override
    public void enqueue(E e) {
        array.addLast(e);
    }

    @Override
    public E dequeue() {
        return array.removeFirst();
    }

    @Override
    public E getFront() {
        return array.getFirst();
    }

    @Override
    public int getSize() {
        return array.getSize();
    }

    @Override
    public boolean isEmpty() {
        return array.isEmpty();
    }

    public int getCapacity() {
        return array.getCapacity();
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append("ArrayQueue: Front [");
        for (int i = 0; i < array.getSize(); i++) {
            sb.append(array.get(i));
            if (i != array.getSize() - 1) sb.append(", ");
        }
        sb.append("] Tail");
        return sb.toString();
    }
}
posted @ 2023-04-10 13:50  lidongdongdong~  阅读(11)  评论(0编辑  收藏  举报