Heap

快速从一堆数据中找到极值

用途：维护一个变化的数据集的最优值。

堆都是完全二叉树逻辑上，堆是一个complete binary tree，即除了最后一层外都是满的，最后一层尽可能往左排。物理上，可以把Heap存在array里，因为它具有complete binary tree的性质。

Left Child Node Index = Parent Node Index *2 +1

Right Child Node Index = Parent Node Index *2 +2

Parent Index = (NodeIndex-1)/2

2. 堆序性

Min-Heap: 每个node的val都比自己孩子节点小（或等于）

Max-Heap：每个node的val都比自己孩子节点大（或等于）

Heap上也有两个基本操作，和stack一样，push和pop

Push/Insert

如果push到minheap的值是一个大的数，可以直接append；

如果push到minheap的值是一个比root还要小的数，sift-up operation；

第一优先级是去满足complete binary tree的性质，先往后面加。

class Heap (object):
    def __init__(self):
        self.array = []

    def sift_up(array, index):
        parent_idx = (index-1)/2
        if parent_idx < 0 or array[index] > array[parent_idx]:
            return
        array[index], array[parent_idx] = array[parent_idx], array[index]
    
        sift_up(array, parent_idx)

    def push(self, val):
        self.array.append(val)
        self.sift_up(self.array, len(self.array)-1)    

arr = [0,1,5,6,8,-1]
sift_up (arr, len(arr)-1)
print(arr)

Time Complexity: O (log n)

Update

如果update value>original, 和下面的min换，如果update value<original,和上面换

Time Complexity: O (log n)

Pop

Pop后为了维护complete binary tree的性质，把最后一个最大的往上提，它和堆顶的换了，然后慢慢Sift-down，接下来update；不能直接update None，不然不能保证最后一层的None在最后还是中间

def sift_down(array, index):
    left = index*2+1
    right = index*2+2
    small = index
    
    if left<len(array) and array[small]>array[left]:
        small = left
    if right<len(array) and array[small]>array[right]:
        small = right
    if small!=index:
        array[small], array[index] = array[index], array[small]
        
        sift_down(array, small)
        
def pop(self):
    res = self.array[0]
    self.array[0], self.array[-1] = self.array[-1], self.array[0]
    self.array.pop()
    self.sift_down(self.array,0)
    return res

Time Complexity: O (log n)

Initialization/Heapify

从Array初始化一个Heap 其实是一个Sift down

def build_heap(arr):
    for i in range(len(arr)/2-1, -1, -1)
        sift_down(arr)

Time Complexity: O(n) 等比数列求和、求极限推导

Smallest K Elements

soln1: sort

O(nlogn)

soln2: quick select/partition

Average Case: O(n+n/2+n/4+n/8+...+1) = O(n)

Worst Case: O(n+ (n-1) + (n-2) + ... ) = O(n^2)

O(kn)

Soln3: 小根堆 Space O(n)

Step 1: Heapify all elements O(n)

Step 2: Call pop() k times to get the k smallest elements. O(klogn)

Time Complexity Total: O(n+klogn)

class Solution(object):
  def kSmallest(self, array, k):
    """
    input: int[] array, int k
    return: int[]
    """
    # write your solution here
    import heapq
    result = []
    heapq.heapify(array)
    for i in range(min(k,len(array))):
      result.append(heapq.heappop(array))
    return result

Soln4: 更适合online 大根堆 Space O(k)

任何时候maxheap里的都是当前的smallest k stream的数据进来，任何时候fixed window with size of k里的元素都是当前的k个最小，之后再有新的stream，都会update when appropriate 流数据找current min之类的也是类似的算法

Step 1: Heapify the first k elements to form a max-heap of size k O(k)

Step 2: Iterate over the remaining n-k elements one by one.

Compare with the largest element of the previous smallest k candidates.

case 1: new element>=top: ignore

case 2: new element<top: update (top ->new element) O((n-k)logk)

Total O(k+(n-k)logk)

def kSmallest2(array, k):
    if not array:
        return array
    if k>=len(array):
        return array
    res = [-elem for elem in array[0:k]]
    heapq.heapify(res)
    for i in range(k,len(array)):
        if array[i] < -res[0]
            heapq.heappop(res)
            heapq.heappush(res, -array[i])
    return [-elem for elem in res]

k~~n指的在一个数量级

Top K Frequent Elements Leetcode 347

Solution 1: 最直接的想法，max heap，全部进heap，取前k个。用-frequency的排序+min heap代替max heap的实现。

Solution 2：用k-size min heap，不断push，把最小的弹出去。注意输出时需要reverse一下。

def topKFrequentPQ(nums, k):
    # 统计频率 O（）
    freq_hash = dict() #0 frequency, 1 number
    for n in numbers:
        if n in freq_hash:
            freq_hash[n]+=1
        else:
            freq_hash[n] =1
    
    # 末尾淘汰 n*2*O(logk)
    heap=[]
    for n, f in freq_hash.items():
        # items()返回的是列表对象，iteritems（）返回的是iterator对象
        heapq.heappush(heap,(f,n))
        if len(heap)>k:
            heapq.heappop(heap)
    
    # 调换顺序 O(klogk)
    while len(heap)!=0:
        ret.append(heapq.heappop(heap)[1])
    ret.reverse()
    
    return ret

在这种方法下不需要每次heapify

Merge K sorted array

Step 1: Create a min heap, put the first element of each array into the heap

Step 2: Each time pop an element from the heap, and then push the next element into the heap.

def mergek(arrays):
    if not arrays:
        return None
    heap = []
    for i in range (len()):
        if len(arrays[i]):
            heap.append((arrays[i][0], i, 0))
        heapq.heapify(heap)
        result=[]
        
        while heap:
            val, index_array, index_element = heapq.heappop(heap)
            result.append(val)
            if index_element+1<len(arrays[index_array]):
                heapq.heappush(heap, (arrays[index_array][index_element+1], 
                index_array, index_element+1))
                
        return result

Time: O(2K+nlogk)

k读入、k来heapify、nlogk来push和popk次

Merge in Stream

def merge_two_streams(A, B):
     
    INVALID = sys.maxsize
    streams = (A, B)
    cache = [(INVALID, 0), (INVALID, 1)]
    for i in [0, 1]:
        try:
            cache[i] = (next(streams[i]), i)
        except StopIteration:
            cache[i] = (INVALID, i) 
    while cache:
        min_val, index = min(cache)
        if min_val == INVALID:
            break
        try:
            cache[index] = (next(streams[index]), index)
        except StopIteration:
            cache[index] = (INVALID, index)
        yield min_val
 
def merge_three_streams(A, B, C):
 
    INVALID = sys.maxsize
    streams = (A, B, C)
    cache = [(INVALID, 0), (INVALID, 1), (INVALID, 2)]
    for i in [0, 1, 2]:
        try:
            cache[i] = (next(streams[i]), i)
        except StopIteration:
            cache[i] = (INVALID, i) 
    while cache:
        min_val, index = min(cache)
        if min_val == INVALID:
            break
        try:
            cache[index] = (next(streams[index]), index)
        except StopIteration:
            cache[index] = (INVALID, index)
        yield min_val
 
from heapq import heapify, heappop, heappush
def merge_k_streams(streams):
     
    cache, k = [], len(streams)
    for i in range(k):
        try:
            cache.append((next(streams[i]),i))
        except StopIteration:
            pass
    heapify(cache)
    while cache:
        min_val, index = heappop(cache)
        try:
            heappush(cache, (next(streams[index]), index))
        except StopIteration:
            pass
        yield min_val
 
def streaming(nums):
    for num in nums: yield num
 
if __name__ == "__main__":
    print([num for num in merge_two_streams(streaming([1,3,5]),streaming([2,4,6]))])
    print([num for num in merge_three_streams(streaming([1,4,7]),streaming([2,5,8]),streaming([3,6,9]))])
    print([num for num in merge_k_streams([streaming(nums) for nums in [[1,3,5,7],[2,4,6,8],[1,4,7,10],[2,5,8,11]]])])

面试题

Linkedin：Can you say something you know about stack and heap?

数据结构的stack：LIFO，

操作系统的stack：函数调用stack存reference，dfs

数据结构的heap：priority queue

操作系统的heap：存object，申请内存来用

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Last updated 4 years ago