我来给出这类题的通用解法，也就是求无序数组的第K大(或第K小)的问题，下面给出两种AC了的方法 import java.util.PriorityQueue;

import java.util.Scanner;

import static java.lang.Integer.parseInt;

import static java.lang.System.in;

public class Main{

public static void main(String[] args) {

Scanner sc = new Scanner(in);

String[] str = sc.nextLine().replace("[", "").replace("]", "").split(",");

int[] data = new int[str.length];

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

data[i] = parseInt(str[i]);

}

System.out.println(findKthNum(data, 3));

System.out.println(findKthNum1(data, 3));

}

//方法1，基于快排思想的partion划分，时间复杂度O(n)，空间复杂度O(1)

public static int findKthNum(int[] data, int k) {

int begin = 0, end = data.length - 1;

int pos = 0;

while (begin <= end) {

pos = partion(data, begin, end);

if (pos == k - 1) {

return data[pos];

} else if (pos > k - 1) {

end = pos - 1;

} else {

begin = pos + 1;

}

}

return -1;

}

private static int partion(int[] data, int begin, int end) {

int temp = data[begin];

while (begin < end) {

while (begin < end && data[end] <= temp) {

end--;

}

swap(data, begin, end);

while (begin < end && data[begin] > temp) {

begin++;

}

swap(data, begin, end);

}

return begin;

}

public static void swap(int[] arr, int i, int j) {

if (arr == null || i >= arr.length || j >= arr.length || i < 0 || j < 0) {

return;

}

int temp = arr[i];

arr[i] = arr[j];

arr[j] = temp;

}

//方法2，建小顶堆，时间复杂度O(n*logk),空间复杂度O(k)

public static int findKthNum1(int data[], int k) {

PriorityQueue heap = new PriorityQueue<>();

for (int item : data) {

if (heap.size() < k) {

heap.add(item);

} else if (item > heap.peek()) {

heap.poll();

heap.add(item);

}

}

return heap.poll();

}

}