<学习记录>Queue源码学习

    xiaoxiao2021-03-25  65

    Queue实现方式(ArrayDeque和LinkedList)ArrayDeque:循环数组elements保存元素

    transient Object[] elements; transient int head; transient int tail;

    构造 public ArrayDeque() { elements = new Object[16]; } public ArrayDeque(int numElements) { allocateElements(numElements); } 调整数组长度: private void allocateElements(int numElements) { int initialCapacity = MIN_INITIAL_CAPACITY; // Find the best power of two to hold elements. // Tests "<=" because arrays aren't kept full. if (numElements >= initialCapacity) { initialCapacity = numElements; initialCapacity |= (initialCapacity >>> 1); initialCapacity |= (initialCapacity >>> 2); initialCapacity |= (initialCapacity >>> 4); initialCapacity |= (initialCapacity >>> 8); initialCapacity |= (initialCapacity >>> 16); initialCapacity++; if (initialCapacity < 0) // Too many elements, must back off initialCapacity >>>= 1;// Good luck allocating 2 ^ 30 elements } elements = new Object[initialCapacity]; }解析:根据特定参数设置初始数组的大小。中间移位和或操作是将initialCapacity左边最高位的1复制到右边,再加1变成2的幂次方。 private void doubleCapacity() { assert head == tail; int p = head; int n = elements.length; int r = n - p; // number of elements to the right of p int newCapacity = n << 1; if (newCapacity < 0) throw new IllegalStateException("Sorry, deque too big"); Object[] a = new Object[newCapacity]; System.arraycopy(elements, p, a, 0, r); System.arraycopy(elements, 0, a, r, p); elements = a; head = 0; tail = n; }解析:这个数组长度扩展,当元素个数增长到等于数组大小时(head==tail),此时将数组大小扩充一倍,再使用arraycopy将元素复制到新的数组。arraycopy(src,srcPos,dest,destPos,length),src指原数组,srcPos指数组复制的开始位置,dest指目的数组,destPos指目的数组的起始位置,length指要复制的长度。 添加 public void addFirst(E e) { if (e == null) throw new NullPointerException(); elements[head = (head - 1) & (elements.length - 1)] = e; if (head == tail) doubleCapacity(); } public void addLast(E e) {         if (e == null)             throw new NullPointerException();         elements[tail] = e;         if ( (tail = (tail + 1) & (elements.length - 1)) == head)             doubleCapacity();     } public boolean offerFirst(E e) {         addFirst(e);         return true;     } public boolean offerLast(E e) {         addLast(e);         return true;     } 添加是将新的元素添加到队头或者队尾,add和offer的功能实现是一样的。因为已经初始化了数组大小,出头元素时head往后移动,如尾元素时tail往后移动,当移动到数组最后时,会将head或者tail循环到数组的前面。所以 head = (head - 1) & (elements.length - 1)的功能相当于 head = (head - 1) %(element.length - 1)由于tail指向的是队尾的后一位,所以先入队尾,再 (tail = (tail + 1) & (elements.length - 1)功能相当于 tail = (tail + 1) % (elements.length - 1)当tail==head时,说明元素已满,所以要扩充数组大小,并使用arraycopy拷贝数组。 移除 public E removeFirst() { E x = pollFirst(); if (x == null) throw new NoSuchElementException(); return x; } public E removeLast() {         E x = pollLast();         if (x == null)             throw new NoSuchElementException();         return x;     } public E pollFirst() {         int h = head;         @SuppressWarnings("unchecked")         E result = (E) elements[h];         // Element is null if deque empty         if (result == null)             return null;         elements[h] = null;     // Must null out slot         head = (h + 1) & (elements.length - 1);         return result;     } public E pollLast() {         int t = (tail - 1) & (elements.length - 1);         @SuppressWarnings("unchecked")         E result = (E) elements[t];         if (result == null)             return null;         elements[t] = null;         tail = t;         return result;     } 解析:poll和remove都是移除元素操作,实现方法一样。出队头元素,将头元素为null,将head移动到下一个位置; head = (h + 1) & (elements.length - 1); 出队尾元素,将最后一个元素为null,tail指向前一个位置。 t = (tail - 1) & (elements.length - 1); 获取元素 public E getFirst() { @SuppressWarnings("unchecked") E result = (E) elements[head]; if (result == null) throw new NoSuchElementException(); return result; } public E getLast() {         @SuppressWarnings("unchecked")         E result = (E) elements[(tail - 1) & (elements.length - 1)];         if (result == null)             throw new NoSuchElementException();         return result;     } public E peekFirst() {         // elements[head] is null if deque empty         return (E) elements[head];     } public E peekLast() {         return (E) elements[(tail - 1) & (elements.length - 1)];     } 解析:peek和get的方法实现一样。主要是获取头元素和尾元素的数组下标。

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