class Test():
    def heap_sort(self, nums):
        i, l = 0, len(nums)
        self.nums = nums
        
        # 构造大顶堆，从非叶子节点开始倒序遍历，因此是l//2 -1 就是最后一个非叶子节点
        for i in range(l//2-1, -1, -1): 
            self.build_heap(i, l-1)
        
        print(nums)
        
        # 上面的循环完成了大顶堆的构造，那么就开始把根节点跟末尾节点交换，然后重新调整大顶堆  
        for j in range(l-1, -1, -1):
            nums[0], nums[j] = nums[j], nums[0]
            self.build_heap(0, j-1)

        return nums
    
    def build_heap(self, i, l): 
        """构建大顶堆"""
        nums = self.nums
        left, right = 2*i+1, 2*i+2 ## 左右子节点的下标
        
        large_index = i 
        
        if left <= l and nums[i] < nums[left]:
            large_index = left

        if right <= l and nums[left] < nums[right]:
            large_index = right
 
        # 通过上面跟左右节点比较后，得出三个元素之间较大的下标，如果较大下表不是父节点的下标，说明交换后需要重新调整大顶堆
        if large_index != i:
            nums[i], nums[large_index] = nums[large_index], nums[i]
            self.build_heap(large_index, l)