通过这道题复习一下python heap操作：
import heapq
python里面heapq用list实现的是最小堆，也就是每次pop（）出来的是最小的数字。有以下method：
q = []
heappush(q, item)
heappop(q) # 如果只想看一下最小的不取出来，则用q[0]即可
heappushpop(q, item) # 先push再pop
heapify(q) # Transform list x into a heap, in-place, in linear time.
heapreplace(q, item) #先pop再push
nlargest(n, q) # return 前n个最大的数作为list
comparator 模版
Screen Shot 2018-12-05 at 12.13.04 PM.png
回到这道题，这道题可以建立一个大小为k的堆，存放最大的k个数，然后新加进来的数再pop掉最小的。
python代码如下：
import heapq
class KthLargest:
def __init__(self, k, nums):
"""
:type k: int
:type nums: List[int]
"""
self.k = k
heapq.heapify(nums)
self.heap = nums
while len(self.heap) > k:
heapq.heappop(self.heap)
def add(self, val):
"""
:type val: int
:rtype: int
"""
if len(self.heap) < self.k:
heapq.heappush(self.heap, val)
else:
heapq.heappushpop(self.heap, val)
return self.heap[0]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)
