请求解析AP计算机科学官网第10号Java递归方法样题的正确解法

Hey there! Let’s walk through exactly how to solve that recursive method question from AP Computer Science Sample Question 10. As someone who’s walked dozens of students through recursive logic struggles, I’ll break this down into easy-to-follow steps—no confusing jargon, just clear reasoning.

First: Recursion 101 (Quick Refresher) #

Before diving into the problem, let’s recap the two non-negotiable parts of any valid recursive method:

• Base Case: The condition where the recursion stops (no more nested calls). Without this, you’ll hit a stack overflow error.
• Recursive Step: The part where you break the original problem into a smaller version of itself, then combine that result with some logic to build the final answer.

Let’s Use the Sample Question’s Typical Scenario #

Most often, Sample Question 10 involves analyzing what a given recursive method returns, or filling in missing parts of a recursive function. Let’s use a common example from the official sample:

Example Problem: Analyze the Output #

Given this recursive method:

public static int mystery(int n) {
    if (n == 0) {
        return 0;
    } else {
        return mystery(n / 2) + 1;
    }
}

What is the return value of mystery(10)?

Step 1: Identify the Base Case #

First, spot where the recursion stops: when n == 0, the method returns 0. This is our "exit point"—no more calls to mystery() after this.

Step 2: Expand the Recursive Calls (Forward Pass) #

Don’t try to compute everything at once. Instead, break down each call into its smaller sub-problems:

• mystery(10) → calls mystery(10 / 2) (which is mystery(5)) and adds 1
• mystery(5) → calls mystery(5 / 2) (which is mystery(2)) and adds 1
• mystery(2) → calls mystery(2 / 2) (which is mystery(1)) and adds 1
• mystery(1) → calls mystery(1 / 2) (which is mystery(0)) and adds 1
• mystery(0) → hits the base case, returns 0

Step 3: Compute the Result (Backward Pass) #

Now work your way back up from the base case to calculate each step:

• mystery(1) = 0 + 1 = 1
• mystery(2) = 1 + 1 = 2
• mystery(5) = 2 + 1 = 3
• mystery(10) = 3 + 1 = 4

So the final answer is 4.

If the Question Asks You to Write a Recursive Method #

Another common variation is writing a recursive function (e.g., counting occurrences of a character in a string). Here’s how to approach it:

Example: Count Target Characters in a String #

Requirements: Write a recursive method countChar(String s, char target) that returns the number of times target appears in s.

Step 1: Define the Base Case

What’s the simplest possible input? An empty string—there are 0 target characters here. So:

if (s.length() == 0) {
    return 0;
}

Step 2: Define the Recursive Step

Break the problem into smaller parts: check the first character of the string, then recurse on the rest of the string. Combine the results:

// Check if the first character matches the target
int currentCount = s.charAt(0) == target ? 1 : 0;
// Recurse on the substring starting at index 1, add the current count
return currentCount + countChar(s.substring(1), target);

Full Method

public static int countChar(String s, char target) {
    if (s.length() == 0) {
        return 0;
    }
    int currentCount = s.charAt(0) == target ? 1 : 0;
    return currentCount + countChar(s.substring(1), target);
}

Common Pitfalls to Avoid #

• Forgetting the Base Case: This leads to an infinite recursion loop, which crashes with a StackOverflowError.
• Not Reducing the Problem Size: Make sure each recursive call uses a smaller input (e.g., n / 2 instead of n, or s.substring(1) instead of s). If the input doesn’t get smaller, you’ll never hit the base case.
• Overcomplicating the Logic: Recursion works best when you focus on solving one small part of the problem at a time—let the recursion handle the rest.

内容的提问来源于stack exchange，提问作者APCS