Frontend Development Learning Path

字数统计: 206字   |   阅读时长: 1min

Frontend Development Learning Path

This learning path provides a structured approach to becoming proficient in frontend development. Follow these modules in sequence for a comprehensive understanding of frontend technologies and best practices.

Module 1: HTML & CSS Fundamentals

Module 2: JavaScript Core

Module 3: Frontend Frameworks

Module 4: Frontend Engineering

Module 5: Advanced Topics

Learning Approach

  1. Theory First: Understand the concepts before writing code
  2. Practice Projects: Build real projects to reinforce learning
  3. Code Review: Get feedback on your code from peers or mentors
  4. Stay Updated: Frontend technologies evolve rapidly - follow blogs, podcasts, and GitHub trends
  • Official documentation (MDN, React docs, etc.)
  • Online platforms: freeCodeCamp, Frontend Masters, Codecademy
  • YouTube channels: Traversy Media, Web Dev Simplified, Academind
  • Books: “Eloquent JavaScript”, “You Don’t Know JS”

Start with Module 1 and progress sequentially. Each topic builds on previous knowledge, creating a solid foundation for frontend development expertise.

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1004.Maximum continuity1Number III Maximum continuity1Number III

字数统计: 297字   |   阅读时长: 1min

Today’s daily question is too difficult，So find the problem by yourself。Today’s question is a hash table+Sliding window algorithm，虽然感觉只用了Sliding window algorithm。

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class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        k_mean = k
        # Used to record how many arrays there are now
        flag = 0
        # Used to record the maximum land number
        max_flag = 0
        for start in range(len(nums)):
            tail = start
            while k >= 0 and tail <= len(nums) - 1:
                if nums[tail] == 1:
                    tail += 1
                    flag += 1
                elif nums[tail] == 0 and k > 0:
                    tail += 1
                    k -= 1
                    flag += 1
                elif nums[tail] == 0 and k == 0:
                    k = k_mean
                    max_flag = max(max_flag, flag)
                    flag = 0
                    break
                if tail == len(nums):
                    max_flag = max(max_flag, flag)
                    flag = 0
                    break
        return max_flag

This is my approach at the beginning，Although the double pointer is used，But there is no flexibility，Very empty feeling，Very dry slide。
the following@Lincoln@Lincoln Big practice，Just as one record of yourself，不作为我的answer发表

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class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        """
        Thinking：1. k=0It can be understood as the problem of the maximum duplication sub -string
             2. If the current window value-In the window1Number <= k: Then expand the window(right+1)
                If the current window value-In the window1Number > k: Swipe the window to the right(left+1)
        method：Hash table + Sliding window
        """
        n = len(nums)
        o_res = 0
        left = right = 0
        while right < n:
            if nums[right]== 1: o_res += 1
            if right-left+1- o_res > k:
                if nums[left]== 1: o_res -= 1 
                left += 1
            right += 1
        return right - left

Look atLincolnBigThinking，very clearly，remember

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