4. If you choose `option 1`, you need to use the `two pointers` algorithm when searching for the other two numbers.

5. For `option 2`, only the `Python` sample code is given. This article focuses on `option 1`.

1. Sort `nums`.

2. Iterate over `nums`.

2. Reverse the order of the words

3. Join the words with a single space

1. Split the string using `split(' ')`

2. Remove empty strings

3. At this time, repeat `node = node.next` on the two linked lists until the same node is found or one of the linked lists has reached the end.

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1. First calculate the number of nodes in the two linked lists A and B. The number of nodes in linked list A is `node_count_a`, and the number of nodes in linked list B is `node_count_b`.

3. When `index == node_count - n`, delete the node by `node.next = node.next.next`.

4. Since the deleted node may be `head`, a virtual node `dummy_node` is used to facilitate unified processing.

1. First find out `node_count`.

2. If `n` has already appeared, it means that the loop has been entered, and `return false`. You can use `Set` to save the `n` that has appeared.

3. Go is the iterative solution, other languages are the recursive solution.

1. Generate a new `n` as the `isHappy(n)` parameter.

<details><summary>Click to view the answer</summary><p>It is `while (current != null)`, because the operation to be performed is `current.next = previous`.</p></details>

1. Traverse all nodes.

For **subarray** problems, you can consider using **Sliding Window Technique**, which is similar to the **Fast & Slow Pointers Approach**.

1. Iterate over the `nums` array, the `index` of the element is named `fastIndex`. Although inconspicuous, this is the most important logic of the *Fast & Slow Pointers Approach*. Please memorize it.

- **High-frequency computation problems**: Such as Fibonacci sequences, factorials, prime number tables, etc., which avoid repetitive calculations by pre-generating lookup tables.

- **Dynamic Programming (DP)**: Pre-computing and storing solutions to sub-problems, e.g., the `knapsack problem` or `shortest path problems`.

1. **Initialize Arrays**:

- Create a `leftProducts` array to store the product of all elements to the left of each element

2. **Two-Phase Calculation**: First compute left products and store in the result, then dynamically compute right products and merge directly

3. **In-Place Operation**: Use only a single variable to dynamically maintain the right product

1. **Initialize Result Array**:

- Create an `answer` array of the same size, initialized with all 1s

1. To solve this problem, you still need to define at least two variables: `current` and `previous`.

2. The loop condition should be `while (current.next != null)` instead of `while (current != null)`, because the operations that need to be performed include `current.next.next`.

1. Traverse all nodes.