703. Kth Largest Element in a Stream（heapq）

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.

Example:

int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3); // returns 4
kthLargest.add(5); // returns 5
kthLargest.add(10); // returns 5
kthLargest.add(9); // returns 8
kthLargest.add(4); // returns 8

Note:
You may assume that nums' length ≥ k-1 and k ≥ 1.

Solution1:(TLE)

class KthLargest:

    def __init__(self, k, nums):
        """
        :type k: int
        :type nums: List[int]
        """
        self.nums = nums
        self.k = k

    def add(self, val):
        """
        :type val: int
        :rtype: int
        """
        self.nums.append(val)
        return sorted(self.nums)[-self.k]

# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)

Solution2:（TLE）

class KthLargest:

    def __init__(self, k, nums):
        """
        :type k: int
        :type nums: List[int]
        """
        self.nums = sorted(nums)[-k:]
        self.k = k
        # print(self.nums)

    def add(self, val):
        """
        :type val: int
        :rtype: int
        """
        if len(self.nums)<k:
            self.nums.append(val)
            self.nums.sort()
        else:
            for i in range(len(self.nums)-1,-1,-1):
                if val>self.nums[i]:
                    for j in range(0,i):
                        self.nums[j] = self.nums[j+1]
                    self.nums[i] = val
                    break
        # print(self.nums)
        return self.nums[0]

这个也TLE过分了

Solution3:

import heapq
class KthLargest(object):

    def __init__(self, k, nums):
        self.pool = nums
        self.k = k
        heapq.heapify(self.pool)
        while len(self.pool) > k:
            heapq.heappop(self.pool)

    def add(self, val):
        if len(self.pool) < self.k:
            heapq.heappush(self.pool, val)
        elif val > self.pool[0]:
            heapq.heapreplace(self.pool, val)
        return self.pool[0]

python的headp为小根堆

heapq.heappush(heap, item)

heapq.heappop(heap)：返回 root 节点，即 heap 中最小的元素。

heapq.heapreplace(heap,item): 替换第一个的值。

heapq.heappushpop(heap, item)：向 heap 中加入 item 元素，并返回 heap 中最小元素。

heapq.heapify(x)：Transform list into a heap, in-place, in O(len(x)) time

heapq.merge(*iterables, key=None, reverse=False)

heapq.nlargest(n, iterable, key=None)：返回可枚举对象中的 n 个最大值，并返回一个结果集 list，key 为对该结果集的操作。

heapq.nsmallest(n, iterable, key=None)：同上相反