从 Python 中的不同类打印？

Posted 2023-02-23

【中文标题】从 Python 中的不同类打印？

【英文标题】：Print from different class in Python?

【发布时间】：2022-01-08 15:12:11

【问题描述】：

我正在复习 Python 的基础知识，并且有两个函数可以使用。解决leet代码问题2的蛮力和字典方法。但是，我试图在使用类时使其工作。我试过把它变成一个大类，现在我把它分成两个类。我不确定如何打印每个函数的结果。请帮忙！

class Solution1:

    def __init__(self, x, y):
        self.x = x
        self.y = y

    # Complexity: O(n^2), very inefficient, especially as the list gets longer
    def twoSum_BF(self, nums, target):
        # don't want to look at the last index (i = index)
        for i in range(len(nums) - 1):
            # don't need to look at indices we've already seen
            # -> i + 1, then go to end of nums
            for j in range(i + 1, len(nums)):
                if nums[i] + nums[j] == target:
                    return [i, j]


# dict method
class Solution2:

    def __init__(self, x, y):
        self.x = x
        self.y = y

    # Complexity is O(n)?
    def twoSum_Dict(self, nums, target):
        seen = 

        # enumerate gives both index and val at same time
        for i, num in enumerate(nums):
            # check whether dict seen contains num
            # needed to add to curr to = target
            # we need tar - curr
            if target - num in seen:
                return [seen[target - num], i]
            elif num not in seen:
                seen[num] = i


x = [2, 7, 11, 15]
y = 9
print("The brute force method returns: ")
BF = (Solution1(x, y))
print(BF)

print("The dictionary method returns: ")
Dict = (Solution2(x, y))
print(Dict)

【问题讨论】：

嗨，欢迎来到堆栈溢出！我的意思是没有冒犯，但是这里有很多小错误可以真正简洁地回答您的问题。我强烈建议您查看一些 class 教程，尝试了解如何使用它们。请参阅w3schools.com/python/python_classes.asp 以获得高级概述，并阅读官方文档以了解更多关于docs.python.org/3/tutorial/classes.html 的确切情况。祝你好运:)

你必须运行这些函数。 BF.twoSum_BF(x, y), Dict.twoSum_Dict(x, y) 得到结果。如果您将x, y 发送到__init__，那么在这些函数中您应该使用self.x、self.y 而不获取参数nums, target

【参考方案1】：

您必须运行函数才能获得结果。

print( BF.twoSum_BF(x, y) )
print( Dict.twoSum_Dict(x, y) )

---

如果您将x,y 发送到__init__，那么您可以在这些函数中使用self.x、self.y 而不是num, target 并在不带参数的情况下运行num, target

print( BF.twoSum_BF() )
print( Dict.twoSum_Dict() )

完整代码：

class Solution1:

    def __init__(self, nums, target):
        self.nums = nums
        self.target = target

    # Complexity: O(n^2), very inefficient, especially as the list gets longer
    def twoSum_BF(self):
        # don't want to look at the last index (i = index)
        for i in range(len(self.nums) - 1):
            # don't need to look at indices we've already seen
            # -> i + 1, then go to end of nums
            for j in range(i + 1, len(self.nums)):
                if self.nums[i] + self.nums[j] == self.target:
                    return [i, j]


class Solution2:

    def __init__(self, nums, target):
        self.nums = nums
        self.target = target

    # Complexity is O(n)?
    def twoSum_Dict(self):
        seen = 

        # enumerate gives both index and val at same time
        for i, num in enumerate(self.nums):
            # check whether dict seen contains num
            # needed to add to curr to = target
            # we need tar - curr
            if self.target - num in seen:
                return [seen[self.target - num], i]
            elif num not in seen:
                seen[num] = i

x = [2, 7, 11, 15]
y = 9

print("The brute force method returns: ")
BF = Solution1(x, y)
print( BF.twoSum_BF() )

print("The dictionary method returns: ")
Dict = Solution2(x, y)
print( Dict.twoSum_Dict() )

【讨论】：

这很有意义，非常感谢！！！