📝 Plan - 三分化

循环一

循环二

⛓️ Technique

通用

• 呼吸能力就是控制姿势的能力
• 意识上和动作的趋势对抗，维持身体的中立
  • 拉背手臂内旋抵抗外旋
  • 推胸手臂外旋抵抗内旋
  • 三头下压手臂外旋抵抗锁骨内旋
  • 飞肩手臂内旋抵抗外旋

Chest

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2Sum

Clarification

The inputs are an integer array nums and an integer target, return the indices of two numbers that add up to target.

Clarifying questions I would ask:

• Are numbers sorted? (Usually no)
• Can there be negative numbers? (Yes)
• Will there be an integer overflow?
• Can I use the same element twice?
• Exactly one solution guaranteed? (Usually yes)
• What should I return if there is no valid answer?

Brute-force

This question is the variation of Stone Game. Let’s start with the brute-force approach. Assume n = piles.length. If we enumerate all possible paths that a player choose either the start or the end stones in the array, players switch between different layers. Each path has a length of n. For each path, we can track the sum scores of two players. At the botom of the decision tree, we compare two player's scores:

Brute-force

Let’s start with the brute-force approach. Assume m = dungeon.length and n = dungeon[0].length. If we enumerate all possible paths from the top-left to the bottom-right cell, at each step we have two choices: move right or move down. Each path has a length of m + n − 1. For each path, we can track the minimum health encountered along the way to determine the minimum initial health required to survive that path. The time complexity is $O(2^{m + n})$. Given the constraints 1 <= m, n <= 200, this brute-force approach is computationally infeasible and will result in a TLE.

蛙泳

Pull-Kick-Glide

1. 分手；

2. 先抬头，自然高肘划水至大臂收紧肋骨两侧，小臂手掌向上贴近身体，口吸气，脚仍然并紧，略微分开；

Intuition

This is an unbounded knapsack problem, meaning there is no upper limit on how many copies of each item (coin) can be used. In general, knapsack-type problems are well suited for dynamic programming. If you are not familiar with DP, you may want to take a look at my website, where I provide a detailed explanation of the core ideas.

Intuition

This is a standard balls-into-bins problem. We make the following assumptions:

1. Each ball must be placed into exactly one bin.
2. The bins are unordered (i.e., permutations of bins are considered equivalent).

Given these assumptions, a backtracking approach is appropriate. However, before discussing the implementation, it is important to recognize that there are two equivalent perspectives from which the backtracking can be formulated. Specifically, we can structure the search by: