求第K大的数~~

已知n个数字各不相同，求其中第k大的数是多少？（1≤k≤n≤10000）

这是一道简单的试题，我们完全可以套用常用的快速排序模型来解决，即对所有数字进行排序，然后取出第k大的数字输出即可，该算法的时间复杂度为O（nlog₂n）

快速排序的基本思想关键在于不斷调整使分治点左边的数不大于（或不小于）分治点，右边的数不小于（或不大于）分治点。

一般快速排序，当找到一个分治点x，序列就以x为分治点，分成左右两部分。

假设每次找到的分治点接近中点，那么，其算法的时间代价大致为：n+n/2+n/4+n/8+…=2n.也就是说改进后的算法的时间复杂度为O(n).

namespace 求第k大的数

{

    class Program

    {

        static int maxn = 10001;//序列长度的上限

        static int[] arr = new int[maxn];//序列

        static void Main(string[] args)

        {

         

            int k, n;//序列长度为n，被寻找的数的大小为k

 

            Console.WriteLine("输入序列的长度");

                n = int.Parse(Console.ReadLine().Trim());

            Console.WriteLine("请输入序列：");

            for (int i = 1; i <= n; i++)

            {

                arr[i] = Convert.ToInt32(Console.ReadLine());

            }

            Console.WriteLine("您要寻找第几大的数？");

            k = int.Parse(Console.ReadLine());

            sort(1,n,k);

       

        }

 

        static void sort(int b_low, int b_high, int k)

        {

            int low = b_low, high = b_high;

            int mid = (low + high) / 2;

            int temp;

            while (low < high)

            {

                while (low < mid)

                {

                    if (arr[low] > arr[mid])

                    {

                        //swap(arr + low, arr + mid);

                        temp = arr[low];

                        arr[low] = arr[mid];

                        arr[mid] = temp;

                        mid = low;

                        low += 1;

                        break;

                    }

                    else

                    {

                        low++;

                    }

                }

                while (mid < high)

                {

                    if (arr[high] < arr[mid])

                    {

                        //swap(arr + high, arr + mid);

                        temp = arr[high];

                        arr[high] = arr[mid];

                        arr[mid] = temp;

                        mid = high;

                        high -= 1;

                        break;

                    }

                    else

                    {

                        high--;

                    }

                }

            }

            if (mid == k)

                Console.WriteLine(arr[mid]);

            else if (mid < k)

            {

                low = mid;

                high = b_high;

                sort(low, high, k);

            }

            else if (mid > k)

            {

                high = mid;

                low = b_low;

                sort(low, high, k);

            }

        }

    }

}