Question

我正在尝试创建一个程序，该程序将允许用户输入一个数字列表，该列表将根据用户的选择通过冒泡，快速，插入或合并排序进行排序。

我正在测试冒泡排序的选择，但是有问题。当从要打印的子例程中取出内容时，变量号将不会打印，其内容是相同的输入列表。

谁能建议一种解决方法，以使气泡排序在用作子例程时正常运行？仅供参考：python的版本是3.6.2，并且不在子例程中时，冒泡排序将起作用。

it 'returns the number provided' do
  # instantiate object that we will test
  subject = RubyOperations::Operations.new

  # we stub method 'gets' and whenever it is called we return string "5"
  allow(subject).to receive(:gets) { "5" } 

  # we call method input with argument 1, the argument 1 is pointless as described in point 1) and can be omitted
  expect(subject.input(1)).to eq(5)
end

Answer 1

您应该使bubble函数return成为结果的numbers列表，以便您的主程序可以打印它。请注意，对于冒泡排序实现是错误的，因为它的上限应涵盖所有项目：

def bubble(numbers):
    import time 
    start = time.time()
    numItems = len(numbers)
    for i in range (0, numItems):
        for j in range (0, numItems - i - 1):
            if numbers [j] > numbers [j + 1]:
                temp = numbers[j]
                numbers[j] = numbers[j + 1]
                numbers[j + 1] = temp
    end = time.time()
    print (end - start)
    return numbers

print ("Bubble, merge, quick and insertion sort visulizer.")
numbers = [int(x) for x in input("Input your list ").split()]
numbers = bubble(numbers)
print(numbers)

尝试在子例程中运行WHILE循环时出错

1 个答案: