python利用列表构造一个优先队列

class priorityQueue:
  def __init__(self):
      self.queue = list()
      self.N = 0
  def leftChild(self, k):
      return k*2+1
  def rightChild(self, k):
      return k*2+2
  def parent(self, k):
      return (k-1)//2
  def less(self, i, j):
      return self.queue[i]<=self.queue[j]
  def exch(self, i, j):
      self.queue[i], self.queue[j] = self.queue[j], self.queue[i]
  #上浮k节点,以最小堆为例子
  def swimMin(self, k):
      while k>0 and self.less(k, self.parent(k)):
          # 当k还没有上浮到根节点，并且k小于它的父节点,则将k上浮到父节点的位置
          self.exch(k, self.parent(k))
          k = self.parent(k)
  #下沉k节点，以最小堆为例子
  def sinkMin(self, k):
      #下沉到堆底，就停止
      while self.leftChild(k)<self.N:
          minEle = self.leftChild(k) #先假设左边最小
          if self.rightChild(k)<self.N and self.less(self.rightChild(k), minEle):
              #如果右孩子更小,更新minEle
              minEle = self.rightChild(k)
          # 如果两个孩子都比父节点大，就不需要下沉了
          if self.less(k, minEle):
              break
          self.exch(minEle, k)
          k = minEle
  def append(self, val):
      self.N += 1
      self.queue.append(val)
      self.swimMin(self.N-1)
  def pop(self):
      if self.N==0:
          return float("nan")
      Min = self.queue[0]
      self.exch(0, self.N-1)
      self.queue.pop()
      self.N -= 1
      self.sinkMin(0)
      return Min    
class Solution:
  def splitArray(self, nums: List[int]) -> int:
      ls = [44,55,36,38,3,7,5]
      priortyQ = priorityQueue()
      for e in ls:
          priorityQ.append(e)#每次往优先队列中加入一个元素
      print(priortyQ.queue)
      print(priortyQ.pop())#弹出一个元素，为优先队列的最小值