堆排序

一、思想

1. 堆的初始化，由最底层逐渐向上一层方向，把最大值向上移动，且$heap(parent) >= max(heap(leftchild), heap(rightchild))$
2. 将最大值移到数组的尾部，将数组尾部的值移动到堆顶。左右孩子的结点各自比较，向上把最大结点的值移动上去。
3. 父节点为n，$leftchild=2^{n}$, $rightchild=2^{n}+1$。（起始为根节点下标为1）

二、一些计算 

1、 满二叉树结点个数$n=2^{i}-1$（等比数列求和$n=frac{a_{0}(1-q^{i})}{1-q}$，复习）。

2、因此最后一个结点的位置设为n，$frac{n}{2}$就是最后一个父节点的位置了。

三、Code

package algorithm;

/**
 * Created by adrian.wu on 2019/2/19.
 */
public class HeapSort {
    /*
    1、Parent * 2 + 1 is left child
    2、Parent * 2 + 2 is right child
     */

    public void headAdjust(int[] array, int parent, int length) {
        int pv = array[parent];

        int left = 2 * parent + 1, right = left + 1;

        while (left < length) {
            if (right < length && array[left] < array[right]) left = right;
            if (pv >= array[left]) break;
            array[parent] = array[left];
            parent = left;
            left = 2 * left + 1;
            right = left + 1;
        }

        array[parent] = pv;
    }

    /*
    1、sort from back to front
    2、
     */
    public void heapSort(int[] array) {
        //create heap, i-1(when root index starting from 0, if 1 that i is 1)
        for (int i = array.length / 2; i >= 0; i--) {
            headAdjust(array, i-1, array.length); 
        }

        //move the head to the end of the list, then recreate the head. 
        for (int i = array.length - 1; i > 0; i--) {
            int temp = array[i];
            array[i] = array[0];
            array[0] = temp;
            headAdjust(array, 0, i);
        }
    }
}