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add min heap
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docs/15_堆与堆排序/min_heap.py

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# python3
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class MinHeap:
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def __init__(self):
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"""
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这里提供一个最小堆实现。如果面试不让用内置的堆非让你自己实现的话,考虑用这个简版的最小堆实现。
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一般只需要实现 heqppop,heappush 两个操作就可以应付面试题了
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parent: (i-1)//2。注意这么写 int((n-1)/2), python3 (n-1)//2当n=0结果是-1而不是0
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left: 2*i+1
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right: 2*i+2
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参考:
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https://favtutor.com/blogs/heap-in-python
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https://runestone.academy/ns/books/published/pythonds/Trees/BinaryHeapImplementation.html
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https://www.askpython.com/python/examples/min-heap
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"""
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self.pq = []
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def min_heapify(self, nums, k):
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"""递归调用,维持最小堆特性"""
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l = 2*k+1 # 左节点位置
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r = 2*k+2 # 右节点
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if l < len(nums) and nums[l] < nums[k]:
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smallest = l
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else:
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smallest = k
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if r < len(nums) and nums[r] < nums[smallest]:
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smallest = r
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if smallest != k:
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nums[k], nums[smallest] = nums[smallest], nums[k]
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self.min_heapify(nums, smallest)
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def heappush(self, num):
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"""列表最后就加入一个元素,之后不断循环调用维持堆特性"""
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self.pq.append(num)
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n = len(self.pq) - 1
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# 注意必须加上n>0。因为 python3 (n-1)//2 当n==0 的时候结果是-1而不是0!
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while n > 0 and self.pq[n] < self.pq[(n-1)//2]: # parent 交换
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self.pq[n], self.pq[(n-1)//2] = self.pq[(n-1)//2], self.pq[n] # swap
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n = (n-1)//2
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def heqppop(self): # 取 pq[0],之后和pq最后一个元素pq[-1]交换之后调用 min_heapify(0)
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minval = self.pq[0]
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last = self.pq[-1]
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self.pq[0] = last
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self.min_heapify(self.pq, 0)
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self.pq.pop()
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return minval
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def heapify(self, nums):
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n = int((len(nums)//2)-1)
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for k in range(n, -1, -1):
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self.min_heapify(nums, k)
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def test_MinHeqp():
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import random
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l = list(range(1, 9))
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random.shuffle(l)
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pq = MinHeap()
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for num in l:
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pq.heappush(num)
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res = []
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for _ in range(len(l)):
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res.append(pq.heqppop()) # 利用 heqppop,heqppush 实现堆排序
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def issorted(l): return all(l[i] <= l[i+1] for i in range(len(l) - 1))
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assert issorted(res)

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