Kth Largest Element in an Array

Leetcode problem description and solution for "Kth Largest Element in an Array".

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Given an integer array nums and an integer k, return the kth largest element in the array.
给定一个整数数组和一个整数 k ，返回数组 nums 中 kth 最大的元素。

Note that it is the kth largest element in the sorted order, not the kth distinct element.
请注意，它是排序顺序中 kth 最大的元素，而不是 kth 非重复元素。

Can you solve it without sorting?
你能在不排序的情况下解决它吗？

Example 1: 示例 1：

Input: nums = [3,2,1,5,6,4], k = 2 Output: 5

Example 2: 示例 2：

Input: nums = [3,2,3,1,2,4,5,5,6], k = 4 Output: 4

Constraints: 约束：

1 <= k <= nums.length <= 105
-104 <= nums[i] <= 104

package org.leetcode;  
import java.util.*;  
public class FindKthLargest {  
    public int findKthLargest(int[] nums, int k) {  
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();  
        for (int num : nums) {  
            minHeap.add(num);  
            if (minHeap.size() > k) {  
                // remove the smallest element  
                minHeap.remove();  
            }  
        }  
        // the kth largest element is at the top of the heap  
        return minHeap.peek();  
  
    }  
  
    public static void main(String[] args){  
        FindKthLargest obj = new FindKthLargest();  
        int[] nums1 = {3,2,1,5,6,4};  
        int k1 = 2;  
        int expect1 = 5;  
        int result1 = obj.findKthLargest(nums1, k1);  
        if (expect1 == result1) {  
            System.out.println("Pass test 1");  
        } else {  
            System.out.println("Fail test 1, expect: " + expect1 + " but got: " + result1);  
        }  
  
        int[] nums2 = {3,2,3,1,2,4,5,5,6};  
        int k2 = 4;  
        int expect2 = 4;  
        int result2 = obj.findKthLargest(nums2, k2);  
        if (expect2 == result2) {  
            System.out.println("Pass test 2");  
        } else {  
            System.out.println("Fail test 2, expect: " + expect2 + " but got: " + result2);  
        }  
    }  
}