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Title 1There is a list structure that contains both ordinary integers and 'lists within lists' (what we often call nested arrays). There's a rule here: the contribution of each integer to the final sum is not equal, but is multiplied by the 'depth' of the integer. For example, the outermost integers have a depth of 1; if you drill down one more layer, the depth of the integers there becomes 2, and so on, with larger depth values the further in you go. Our task is to figure out the sum of all those integers, and each integer has to be 'weighted' according to the depth at which it lies.

Problem Solving Ideas:

Doesn't this question sound a bit like playing a treasure hunt? The value of each treasure (integer) depends on how deep it is hidden. We have to dig up each treasure, multiply it by the depth of the layer it is in, and then add up the value of all the "weighted treasures".

Core Ideas:

1. Iteration: We have to find all the elements in this nested list one by one.
2. Deep Tracking: Here's the key! Whenever we enter a sub-list, the depth has to be increased by 1. Whenever we finish processing a sub-list and come out of it, the depth has to be decreased by 1. This way, we always know where the currently processed integer is.
3. Cumulative: Find the integer, multiply it by the current depth, and add it to the sum.

The solution process (in Python, for example, because it is intuitive to express nested structures):

Imagine we hold a treasure hunting map in our hands with this structure drawn on it: [[1,1],2,[1,1]]

First solution: recursive bigrama (most common and elegant)

Recursion is like a scout that knows how to go deeper in layers and also return to the previous layer exactly.

• Function Design: We need a function called for example calculate_weighted_sum(nested_list, depth).

  • nested_list: the current list to be processed (may be top level or sub-list).
  • depth: depth of the current list.

• Recursive Logic:

  1. Initialization: Set a total_sum = 0 to accumulate the sum.
  2. Iterates through the current list: Loop over each element in nested_list.
     
     • If element is an integer (int):
       
       • Congratulations, you've found the treasure! Its value is element * depth.
       • Add this value to total_sum.
     • If element is a list:
       
       • Aha! Another treasure hole! That means we're going in deeper.
       • Call ourselves, that is, recursively call calculate_weighted_sum(element, depth + 1).
       • Add the result of this recursive call (the weighted sum of the sublists) to total_sum.
  3. Return: When the loop ends, indicating that all elements of the current list have been processed, return total_sum.

• Initial call: The first time you call it, you start at the top, of course, with a depth of 1. So you would call it like this: calculate_weighted_sum(your_main_nested_list, 1).

Take a chestnut ([[1,1],2,[1,1]]):

1. Call: calculate_weighted_sum([[1,1],2,[1,1]], 1)
   
   • total_sum = 0
   • Processing [1,1] (first element). Found it to be a list!
     
     • Recursive call: calculate_weighted_sum([1,1], 2)
       
       • sub_total_sum = 0
       • Treatment 1. Integer, 1 * 2 = 2. sub_total_sum becomes 2.
       • Treatment 1. integer, 1 * 2 = 2. sub_total_sum becomes 2 + 2 = 4.
       • Return 4
     • total_sum becomes 0 + 4 = 4.
   • Treatment 2 (second element). It was found to be a whole number!
     
     • 2 * 1 = 2.
     • total_sum becomes 4 + 2 = 6.
   • Processing [1,1] (third element). Found it to be a list.
     
     • Recursive call: calculate_weighted_sum([1,1], 2)
       
       • (As above, this call also returns 4)
     • total_sum becomes 6 + 4 = 10.
   • Back to 10

Recursion is used just right at this point; it drills itself in and out layer by layer, keeping track of depth perfectly.

Second solution: iterative macromethod (with stacks or queues)

While recursion is beautiful, sometimes we may prefer iteration because it controls the flow more intuitively and avoids the recursion depth limitations of some languages.

• Data structure: We need a stack (or a queue, depending on the traversal order, it's more natural to use a stack here) to hold the elements to be processed and their depths. Stacks are characterized as "last in, first out", which is ideal for this kind of "drill-in-pop-out" scenario.

• Element in the stack: Instead of a single element, the stack holds a "tuple" or "pair", such as (element, current_depth).

• Iterative Logic:

  1. Initialization: total_sum = 0.
  2. Prepare the stack: Presses all the elements of the top-level list and their initial depth 1 onto the stack. Note that we usually press them in reverse order so that the loop will be processed in positive order when it pops up.
  3. Loop: It keeps looping as long as the stack is not empty:
     
     • Pop up: Pop a (current_element, current_depth) from the top of the stack.
     • Type of judgment:
       
       • If current_element is an integer:
         
         • total_sum += current_element * current_depth.
       • If current_element is a list:
         
         • This means that we have to "enter" this sub-list.
         • Press all the elements in this sublist, along with the new depth current_depth + 1, onto the stack again. Again, press them in reverse order to maintain processing order.

• When it ends: When the stack is empty, all elements have been processed and total_sum is the final result.

Take a chestnut ([[1,1],2,[1,1]]):

1. Initialization: total_sum = 0

2. Stack (initial):

   • stack = []
   • stack.append(([1,1], 1)) (assuming we can just press the list as a whole)
   • stack.append((2, 1))
   • stack.append(([1,1], 1))
   • Now stack = [([1,1], 1), (2, 1), ([1,1], 1)] (note that here, for simplicity, we are assuming that the list is pressed as a whole; in practice it may be necessary to press the elements of the top-level list one by one first)

   For a more accurate simulation, we should initially look like this:

   • stack = []
   • Iterate over [[1,1],2,[1,1]]:
     
     • Press ([1,1], 1) into the
     • Press (2, 1) into
     • Press ([1,1], 1) into the
   • This way, when we pop it, it will process ([1,1], 1), then (2, 1), and finally ([1,1], 1).

   A more rigorous initial press: stack = [(element, 1) for element in nested_list[::-1]] (pressed in reverse order to ensure positive order when popped)

3. Loop:

   • Popup ([1,1], 1): It's a list!
     
     • Press (1, 2) onto the stack
     • Press (1, 2) onto the stack
     • stack = [(2, 1), ([1,1], 1), (1, 2), (1, 2)]
   • Pop up (1, 2): This is an integer! total_sum += 1 * 2 = 2.
   • Pop up (1, 2): This is an integer! total_sum += 1 * 2 = 4.
   • Eject (2, 1): This is an integer! total_sum += 2 * 1 = 6.
   • Popup ([1,1], 1): It's a list!
     
     • Press (1, 2) onto the stack
     • Press (1, 2) onto the stack
     • stack = [(1, 2), (1, 2)]
   • Eject (1, 2): This is an integer! total_sum += 1 * 2 = 8.
   • Eject (1, 2): This is an integer! total_sum += 1 * 2 = 10.
   • The stack is empty and the loop ends.

The final result is also 10.

Title 2: "We've got a string. $Q$, which consists only of parentheses () and center brackets []. To determine whether such a string is 'correct', there are several rules:

(a) If the string is empty, it is 'correct'. (b) If the string $A$ and strings $B$ are 'correct', then the string formed by stitching them together $AB$ is also 'correct'. (c) If the string $A$ is 'correct,' then the string formed after the parentheses (A) or the center bracket [A] are placed around it is also 'correct.'"

Problem Solving Ideas:

This definition actually describes a recursive structure, very similar to the way we define "legal arithmetic expressions" or "legal XML/HTML tags".

• (a) The empty string is correct: This rule is very important, it is the "base case" of recursion. Imagine if you don't have any parentheses, it's naturally "correct" because it doesn't contain any "wrong" matches. It's like building a house, where an empty lot is legal and you can start building on it.

• (b) Splicing of correct strings: This rule explains the correctness of the "tandem". If $A$ is a complete, well-matched structure that $B$ is also a complete, well-matched structure, then put them next to each other and of course they still match well. For example, if () is correct, the$[]$ is also correct, then ()[] is correct. It's like if you build two separate legal houses, put them next to each other, and they're still two legal houses.

• (c) The closure of the correct string: This rule explains the correctness of "nesting". If $A$ itself is a correct, matching structure, then wrap it in its entirety inside a pair of matching parentheses () or [] and the new structure is still correct. For example, if () is correct, then both (()) and [()] are correct. It's like if you add another legal fence around the outside of a legal house, the whole structure is still legal.

These three rules together define the validity of "bracket matching". It covers all possible "legal" combinations:

• No content (empty).
• Parallel arrangement (splicing).
• Layers are nested (closed).

With these few simple rules, we can determine that strings like ([]), ()[[()]] are correct, and strings like ([)], (() are incorrect. This has a wide range of applications in areas such as compiler design, data structures (the application of stacks to check for bracket matching), and so on.