Kth Largest Element in an Array leetcode

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.
用quick selecte的方法

//pivot为left
public class Solution {
    public int findKthLargest(int[] nums, int k) {
        return find(nums, 0, nums.length - 1, nums.length - k);
    }
    public int find (int[] nums, int i, int j, int k) {
        int left = i;
        int right = j;
        int pivot = nums[left];
        while (left < right) {//为< 不是<=
            while (left < right && nums[right] > pivot) {//必须先right--
                right--;
            }
            while (left < right && nums[left] <= pivot) {
                left++;
            }
            
            swap(nums, left, right);
            
        }
        swap(nums,right, i);
        if (left == k) {
            return nums[k];
        } else if (left < k) {
            return find(nums, left + 1, j, k);
        } else {
            return find(nums, i, left - 1, k);
        }
    }
    public void swap(int[] nums, int i, int j) {
        int tem = nums[i];
        nums[i] = nums[j];
        nums[j] = tem;
    }
}
//pivot为right
public class Solution {
    public int findKthLargest(int[] nums, int k) {
        return find(nums, 0, nums.length - 1, nums.length - k);
    }
    public int find (int[] nums, int i, int j, int k) {
        int left = i;
        int right = j;
        int pivot = nums[right];
        while (left < right) {
            while (left < right && nums[left] < pivot) {//右pivot时候先左边 反之亦然
                left++;
            }
            while (left < right && nums[right] >= pivot) {
                right--;
            }
            if (left < right) {//此判定可有可无, 因为不会出现left> right情况
                swap(nums, left, right);
            }
            
        }
        swap(nums,left, j);
        if (left == k) {
            return nums[k];
        } else if (left < k) {
            return find(nums, left + 1, j, k);
        } else {
            return find(nums, i, left - 1, k);
        }
    }
    public void swap(int[] nums, int i, int j) {
        int tem = nums[i];
        nums[i] = nums[j];
        nums[j] = tem;
    }
}

Priority Queue的解法

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        PriorityQueue<Integer> queue = new PriorityQueue<Integer>();
        for (int i : nums) {
            queue.offer(i);
        }
        for (int i = 0; i < nums.length - k ; i++) {
            queue.poll();
        }
        return queue.peek();
    }
}