Java字符串匹配任务求助：基于优先级的字符匹配类型判定逻辑实现

Got it, let's pivot away from that backspace code and tackle this string matching task properly. The key here is following the priority rules strictly: P first, then S, then I, and making sure we track which characters from A are still available for matches.

Core Logic Breakdown #

1. First pass: Mark all P matches: These are the highest priority. Every time A[i] equals B[i], we mark that position as P and decrement the count of that character in A (since it's now used).
2. Second pass: Mark S matches: For each position that isn't already P, check if B[i] exists in the remaining unused characters of A. If yes, mark it as S and decrement the count again.
3. Remaining positions stay I: Any spot not marked P or S is innocent.

Using a frequency array (for 26 uppercase letters) is perfect here—it's O(1) space and super fast, even for the maximum input size of 1e6 characters.

Pseudocode #

// Initialize a count array for A's characters (26 for each uppercase letter)
countA = array of 26 zeros

// First, count all characters in string A
for each character c in A:
    countA[c - 'A'] += 1

// Initialize result array with 'I' (default)
result = array of 'I's with length N

// Step 1: Process all P matches
for i from 0 to N-1:
    if A[i] == B[i]:
        result[i] = 'P'
        countA[A[i] - 'A'] -= 1  // Mark this character as used for P

// Step 2: Process all S matches
for i from 0 to N-1:
    if result[i] == 'P':
        continue  // Skip already matched positions
    bChar = B[i]
    charIndex = bChar - 'A'
    if countA[charIndex] > 0:
        result[i] = 'S'
        countA[charIndex] -= 1  // Mark this character as used for S

// Convert result array to string and output
print(result as string)

Java Implementation #

This code is optimized for large input sizes (1e6 characters) with linear time complexity O(N) and constant space O(1):

import java.util.Scanner;

public class StringMatchClassifier {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String A = scanner.nextLine().trim();
        String B = scanner.nextLine().trim();
        int n = A.length();
        
        // Count frequency of each character in A
        int[] charCountA = new int[26];
        for (char c : A.toCharArray()) {
            charCountA[c - 'A']++;
        }
        
        char[] result = new char[n];
        // Step 1: Mark all P matches first
        for (int i = 0; i < n; i++) {
            char aChar = A.charAt(i);
            char bChar = B.charAt(i);
            if (aChar == bChar) {
                result[i] = 'P';
                charCountA[aChar - 'A']--;
            } else {
                result[i] = 'I'; // Default to innocent
            }
        }
        
        // Step 2: Mark S matches for remaining positions
        for (int i = 0; i < n; i++) {
            if (result[i] == 'P') {
                continue;
            }
            char bChar = B.charAt(i);
            int idx = bChar - 'A';
            if (charCountA[idx] > 0) {
                result[i] = 'S';
                charCountA[idx]--;
            }
            // Leave as 'I' if no available character in A
        }
        
        System.out.println(new String(result));
    }
}

How This Works With the Examples #

Let's take Example 1 to verify:

• Input A: CLOUD, B: CUPID
• Initial countA: C=1, L=1, O=1, U=1, D=1
• After P pass: positions 0 (C=C) and 4 (D=D) are marked P. countA becomes C=0, L=1, O=1, U=1, D=0
• After S pass: position 1 (B's U) has countA[U] = 1, so mark as S. countA[U] becomes 0. Positions 2 (P) and 3 (I) have no matches in A, so stay I.
• Final output: PSIIP which matches the example.

This approach efficiently handles all edge cases, including repeated characters in A and B, while adhering strictly to the priority rules.

内容的提问来源于stack exchange，提问作者Pete Alexandrovich