第k大的元素

析：借助快速排序的思想

import java.util.Scanner;

public class Main4 {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        String numStr=sc.nextLine();
        String[] nums = numStr.split(" ");
        int[] arr = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            arr[i]= Integer.valueOf(nums[i]);
        }
        int k=sc.nextInt();
        System.out.println(kthLargestElement(k,arr));

    }

    public static int kthLargestElement(int k, int[] nums) {
        if(null == nums ||k <= 0 || nums.length < k) return Integer.MIN_VALUE;
        kthMax(nums, 0, nums.length-1, k);
        return nums[k-1];
    }
    //  快速排序的近似写法
    public static void kthMax(int[] nums, int left, int right, int k) {
        int mid = partition(nums, left, right);
        if(mid == k-1) return;
        if(mid > k-1)
            kthMax(nums, left, mid-1, k);
        else
            kthMax(nums, mid+1, right, k);
    }

    public static int partition(int[] nums, int left, int right) {
        int temp = nums[left];
        while(left < right) {
            while(left < right && nums[right] <= temp)
                right--;
            nums[left] = nums[right];
            while(left < right && nums[left] > temp)
                left++;
            nums[right] = nums[left];
        }
        nums[left] = temp;
        return left;
    }
}