洛谷P1923 【深基9.例4】求第 k 小的数（java）

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.Arrays;
import java.util.Scanner;
//输入n个数字ai，输出这些数字的第k小的数。最小的数是第0小。
public class Main {

    public static void main(String[] args) throws IOException {
        StreamTokenizer sc = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        sc.nextToken();
        int n = (int) sc.nval;
        sc.nextToken();
        int k = (int) sc.nval;
//        Scanner sc = new Scanner(System.in);
//        int n = sc.nextInt();//n个数
//        int k = sc.nextInt();//第k小的数
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            sc.nextToken();
            a[i] = (int) sc.nval;
        }

        for (int i = 0; i < a.length; i++) {
            for (int j = i; j < a.length; j++) {

            }
        }
        //Quick_sort(a, 0, a.length - 1);
        //System.out.println(a[k]);o（nlogn）
        int i = find(a, 0, a.length - 1, k);

    }
     public static int find (int[] a,int left, int right, int k)
    {	//在数组a的第left到right中寻找第k小的数
        int tem=Partition(a,left,right);
        if(k==tem)
            System.out.println(a[k]);
        else if(k-1= pivot)
                high--;
            a[low] = a[high];
            while (low < high && a[low] <= pivot) low++;
            a[high] = a[low];
        }
            a[low] = pivot;
            return low;//return high
    }
}