215. Kth Largest Element in an Array

Description

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.

Note: You may assume k is always valid, 1 ≤ k ≤ array's length.

Constraints

Approach

Examples

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

Output: 5

Solutions

/**
 * Time complexity : O(NlogN)
 * Space complexity : O(1)
 */

class Solution {
    public int findKthLargest(int[] nums, int k) {
        Arrays.sort(nums);
        return nums[nums.length-k];
    }
}

/**
 * Time complexity : O(N)
 * Space complexity : O(N)
 */

class Solution {
    public int findKthLargest(int[] nums, int k) {
        PriorityQueue<Integer> queue = new PriorityQueue(k);
        for(int n: nums) {
            queue.offer(n);
            if(queue.size() > k) {
                queue.poll();
            }
        }
        return queue.peek();
    }
}

/**
 * Time complexity : O(N) in the average case, O(N^2)in the worst case.
 * Space complexity : O(1)
 */

class Solution {
    public int findKthLargest(int[] nums, int k) {
        return qs(nums, nums.length-k+1, 0, nums.length-1);
    }
    
    private int qs(int[] nums, int k, int start, int end) {
        int left = start,
            right = end;
        int pivot = nums[end];
        
        while(true) {
            while(left < right && nums[left] < pivot) {
                left++;
            }
            while(right > left && nums[right] >= pivot) {
                right--;
            }
            
            if(left == right) break;
            
            swap(nums, left, right);
        }
        
        swap(nums, left, end);
        
        if(k == left+1) {
            return pivot;
        } else if(k < left+1) {
            return qs(nums, k, start, left-1);
        } else {
            return qs(nums, k, left+1, end);
        }
    }
    
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}

Follow up

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