215 Kth Largest Element in an Array

215. Kth Largest Element in an Array

1. Question

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.

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

2. Implementation

(1) Quick Select

class Solution {
    public int findKthLargest(int[] nums, int k) {
        return getKthNum(nums, 0, nums.length - 1, k - 1);
    }

    public int getKthNum(int[] nums, int start, int end, int k) {
        int pivot = partition(nums, start, end);

        if (pivot < k) {
            return getKthNum(nums, pivot + 1, end, k);
        }
        else if (pivot > k) {
            return getKthNum(nums, start, pivot - 1, k);
        }
        else {
            return nums[pivot];
        }
    }

    public int partition(int[] nums, int start, int end) {
        int mid = start + (end - start) / 2;

        swap(nums, mid, end);

        for (int i = start; i < end; i++) {
            if (nums[i] > nums[end]) {
                swap(nums, i, start);
                ++start;
            }
        }

        swap(nums, start, end);
        return start;
    }

    public void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

(2) Heap

class Solution {
    public int findKthLargest(int[] nums, int k) {
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();

        for (int num : nums) {
            minHeap.add(num);

            if (minHeap.size() > k) {
                minHeap.remove();
            }
        }
        return minHeap.peek();
    }
}

3. Time & Space Complexity

Quick Select: 时间复杂度O(n), 空间复杂度O(1)

Heap: 时间复杂度O(n), 空间复杂度O(k)