算法导论Java实现-堆排序(6.4章节)

package lhz.algorithm.chapter.six;
/**
 * “堆排序”，《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(苦逼coder)  * 本文地址：http://mushiqianmeng.blog.51cto.com/3970029/738164  */
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 "discard" 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 + " ");
        }
    }
}