用heap解，

方法1. 維護一個 size = k 的最小堆。當前元如果大於堆頂的元素，那麽說明堆頂的元素肯定小於kth largest element。所以replace他。

 1 class Solution(object):
 2     def findKthLargest(self, nums, k):
 3         """
 4         :type nums: List[int]
 5         :type k: int
 6         :rtype: int
 7         """
 8         heap = []
 9         res = 0
 
10         
11         for i in range(len(nums)):
12             if i < k:
13                 heapq.heappush(heap, nums[i])
14             else:
15                 if heap[0] < nums[i]:
16                     heapq.heapreplace(heap, nums[i])
17             
18  
19         return heap[0]

或者維護一個-nums的最小堆，從heap pop出第k個元素。那麽這個數就是 -nums的第k小元素，也就是nums的第k大元素。

class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        heap = []
        res = 0
        
        for i in range(len(nums)):
            heapq.heappush(heap, -nums[i])
        
        for
 
 j in range(k):
            res = -heapq.heappop(heap)
            
        return res

215. Kth Largest Element in an Array