堆排序 - Java实现

package lhz.algorithm.chapter.six;
/**
 * "排序"，《算法导论》6.4章节
 * 利用之前实现的构建MaxHeap和MaxHeapify算法完成排序。
 * 伪代码：
 * HEAPSORT(A)
 * 1 BUILD-MAX-HEAP(A)
 * 2 for i <- length[A] downto 2
 * 3 do exchange A[1] <-> A[i]
 * 4 heap-size[A] <- heap-size[A] - 1
 * 5 MAX-HEAPIFY(A, 1)
 * @author lihzh(OneCoder)
 * @OneCoder-Blog http://www.coderli.com
 */
public class HeapSort {
	
	private static int[] input = new int[] {16, 14, 10, 8, 7, 9, 3, 2, 4, 1};

	public static void main(String[] args) {
		//堆排序
		heapSort();
		//打印数组
		printArray();
	}
	
	/**
	 * 堆排序，《算法导论》原文摘要：
	 * The heapsort algorithm starts by using BUILD-MAX-HEAP to build a max-heap on the input
	 * array A[1 - n], where n = length[A]. Since the maximum element of the array is stored at the
	 * root A[1], it can be put into its correct final position by exchanging it with A[n]. 
	 * If we now &quot;discard&quot; node n from the heap (by decrementing heap-size[A]), we observe that 
	 * A[1 , (n -1)] can easily be made into a max-heap. The children of the root remain max-heaps, 
	 * but the new root element may violate the max-heap property. All that is needed to restore 
	 * the maxheap property, however, is one call to MAX-HEAPIFY(A, 1), which leaves a max-heap 
	 * in A[1, (n - 1)]. The heapsort algorithm then repeats this process for the max-heap of size 
	 * n - 1 down to a heap of size 2.
	 * 复杂度：
	 * 由之前分析可知，buildMaxHeap复杂度为O(n lg n)，运行一次。
	 * maxHeapify的复杂度为O(lg n)，运行n-1次。
	 * 综上，复杂度为O(n lg n)。
	 */
	private static void heapSort() {
		int length = input.length;
		//构造max-heap
		buildMaxHeap(input, length);//交换位置
		for (int i = length - 1; i > 0; i--) {
			int temp = input[i];
			input[i] = input[0];
			input[0] = temp;
			maxHeapify(input, 1, i);
		}
	}

	private static void buildMaxHeap(int[] array, int heapSize) {
		for (int i = heapSize / 2; i > 0; i--) {
			maxHeapify(array, i, heapSize);
		}
	}
	
	private static void maxHeapify(int[] array, int index, int heapSize) {
		int l = index * 2;
		int r = l + 1;
		int largest;
		//如果左叶子节点索引小于堆大小，比较当前值和左叶子节点的值，取值大的索引值
		if (l <= heapSize && array[l-1] > array[index-1]) {
			largest = l;
		} else {
			largest = index;
		}
		//如果右叶子节点索引小于堆大小，比较右叶子节点和之前比较得出的较大值，取大的索引值
		if (r <= heapSize && array[r-1] > array[largest-1]) {
			largest = r;
		}
		//交换位置，并继续递归调用该方法调整位置。
		if (largest != index) {
			int temp = array[index-1];
			array[index-1] = array[largest-1];
			array[largest-1] = temp;
			maxHeapify(array, largest, heapSize);
		}
	}
	
	private static void printArray() {
		for (int i : input) {
			System.out.print(i + " ");
		}
	}
}