快排以及基于快排思想的topk 一锅端demo

算法不好，补补基本的排序算法，弄懂了快排，发现topK问题中也能用快排的思想所以写了一个demo

填坑解释法解释快排很形象，我是看这篇看懂快排的,这里推荐一下
http://blog.csdn.net/chengqiaoyahaixuyush/article/details/38273893

以下是快排和基于快排思想的topK问题的实践demo

public class TopK_Quick {

    public static void main(String args[]) {
        int k = 3;
        int a[] = {2, 0, 5, 4, 3, 1};
        if (k > 0 && k <= a.length - 1) {
            selectK(a, 0, a.length - 1, k);
            display(a, k);

            //因为执行完selectK 拿到的top K是无序的，这里对top K 个元素再排序
            int[] b = new int[k];
            System.arraycopy(a, 0, b, 0, k);
            QuickSort(b, 0, b.length - 1);
            display(b, k);
        } else {
            System.out.println("Are You Kidding Me?");
        }
    }

    public static int PartitionLow(int a[], int low, int high) {
        int pivokey = a[low];
        while (low < high) {
            while (low < high && a[high] >= pivokey) --high;
            a[low] = a[high];
            while (low < high && a[low] <= pivokey) ++low;
            a[high] = a[low];
        }
        a[low] = pivokey;
        return low;
    }

    public static int PartitionHigh(int a[], int low, int high) {
        int pivokey = a[low];
        while (low < high) {
            while (low < high && a[high] <= pivokey) --high;
            a[low] = a[high];
            while (low < high && a[low] > pivokey) ++low;
            a[high] = a[low];
        }
        a[low] = pivokey;
        return low;
    }

    public static void QuickSort(int a[], int low, int high) {
        if (low >= high || a.length <= 1) {
            return;
        }
        int i = low;
        int j = high;
        int x = a[i];
        while (i < j) {
            while (i < j && a[j] >= x) j--;
            if (i < j) {
                a[i++] = a[j];
            }

            while (i < j && a[i] < x) i++;
            if (i < j) {
                a[j--] = a[i];
            }
        }
        a[i] = x;
        QuickSort(a, low, i - 1);
        QuickSort(a, i + 1, high);
    }

    public static void display(int[] a, int k) {
        for (int i = 0; i < k; i++) {
            System.out.print(a[i] + " ");
        }
        System.out.println("");
    }

    public static int selectK(int a[], int start, int end, int k) {
        int index = 0;
        if (start < end) {
            //获取最大K个数
            index = PartitionHigh(a, start, end);

            //获取最小K个数
            // index = PartitionLow(a, start, end);

            if (index == k)//正好找到第k大的数
            {
                index = k;
            } else if (index < k)//还要从index的右边找k-index个数
            {
                index = selectK(a, index + 1, end, k - index);
            } else if (index > k)//k个数都在Index的左边
            {
                index = selectK(a, start, index - 1, k);
            }
        }
        return index;

    }
}