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;    }}