215. Kth Largest Element in an Array

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.

Example 1:

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

Example 2:

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

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

 

Approach #1: C++. [priority_queue]

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        priority_queue<int, vector<int>, greater<int>> pq;
        
        for (int i = 0; i < nums.size(); ++i) {
            if (pq.size() <= k) pq.push(nums[i]);
            else pq.pop(), pq.push(nums[i]);
        }
        if (pq.size() > k) pq.pop();
        return pq.top();
    }
};

　　

Approach #2: Java. [quick select]

class Solution {
    public int findKthLargest(int[] nums, int k) {
        int n = nums.length;
        int p = quickSelect(nums, 0, n-1, n-k+1);
        return nums[p];
    }
    
    int quickSelect(int[] a, int lo, int hi, int k) {
        int i = lo, j = hi, pivot = a[hi];
        while (i < j) {
            if (a[i++] > pivot) swap(a, --i, --j);
        }
        
        swap(a, i, hi);
        
        int m = i - lo + 1;
        
        if (m == k) return i;
        else if (m > k) return quickSelect(a, lo, i-1, k);
        else return quickSelect(a, i+1, hi, k-m);
    }
    
    void swap(int[] a, int i, int j) {
        int tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }
}

Analysis;

In this case, we swap the elements to make the array ordered. If the number of minimum elements in front of the array equal to k, then return the position at the array. Otherwise we divide the array with the pivot, if m(the elements of minimum numbers in front of the array) bigger the k, we can find the right position in the smaller numbers partion, Otherwise finding in the bigger partion.

 

Approach #3: Python.

import heapq
class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        min_heap = nums[:k]
        heapq.heapify(min_heap)
        for i in range(k, len(nums)):
            if nums[i] > min_heap[0]:
                heapq.heappop(min_heap)
                heapq.heappush(min_heap, nums[i])
        return min_heap[0]