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ARTS #104

Algorithm

本周选择的算法题是：Min Stack。

规则

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

• MinStack() initializes the stack object.
• void push(val) pushes the element val onto the stack.
• void pop() removes the element on the top of the stack.
• int top() gets the top element of the stack.
• int getMin() retrieves the minimum element in the stack.

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

Constraints:

• -231 <= val <= 231 - 1
• Methods pop, top and getMin operations will always be called on non-empty stacks.
• At most 3 * 104 calls will be made to push, pop, top, and getMin.

Solution

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        import math

        self.stack = []
        self.min_stack = [math.inf]

    def push(self, val: int) -> None:
        self.stack.append(val)
        self.min_stack.append(min(val, self.min_stack[-1]))

    def pop(self) -> None:
        self.stack.pop()
        self.min_stack.pop()

    def top(self) -> int:
        return self.stack[-1]

    def getMin(self) -> int:
        return self.min_stack[-1]

Review

Swift 5: Why Is Protocol-Oriented Programming Essential

一篇介绍什么是面向协议编程的文章，作者用清晰而易懂的方法讲解了如何将一个 God Class 变得更干净，对大型、复杂工程非常有帮助。

在实际开发过程中，特别容易形成多种角色混在一起的 God Class，比如这篇 抖音 iOS 最复杂功能的重构之路 – 播放器交互区重构实践 文章中提到的，一个 VC 就有 1.8 万行代码，重构的难度、风险可想而知，但是不重构又有维护性差、可读性差、扩展性差等问题，业务交付越来越困难，重构只是时间早晚而已。

面向协议的编程方法在面对复杂场景时还是有独特优势的。

Tip

发现了一个 macOS 下识别截屏文案的应用：TextSniper，难得的是对中文的支持也很好。

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