java如何找到给定字符串中的每个回文子字符串，并在不到一秒钟的时间内返回一个出现值

Question

我一直在努力解决以下问题：

You are given a string of lower-case Latin letters. Let us define a substring's "occurrence value" as the number of the substring occurrences in the string multiplied by the length of the substring. For a given string find the largest occurrence value of palindromic substrings.

我的代码工作得很好，但是，我需要在不到一秒钟的时间内输入多达30万个字符的解决方案。我的代码如下：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;

public class Palindrome {

public static void main(String[] args) {
    // initiate a scanner
    Scanner in = new Scanner(System.in);
    String pal = in.nextLine();
    getAllPalindrome(pal);

}

/**
 * checks if the given string is a palindrome
 * 
 * @param pal
 * @return
 */
public static boolean checkPalindrome(String pal) {
    for (int i = 0; i < pal.length() / 2; i++) {
        if (pal.charAt(i) != pal.charAt(pal.length() - 1 - i)) {
            return false;
        }

    }
    return true;
}

/**
 * gets all palindromes
 * 
 * @param pal
 */
public static void getAllPalindrome(String pal) {
    // initiate variables
    ArrayList<String> pals = new ArrayList<String>();
    int count = 0;
    // add all palindromes to an arraylist
    for (int i = 0; i < pal.length(); i++) {
        for (int j = i; j < pal.length(); j++) {
            if (checkPalindrome(pal.substring(i, j + 1))) {
                pals.add(pal.substring(i, j + 1));
            }
        }
    }

    int[] counts = new int[pals.size()];
    for (int i = 0; i < pals.size(); i++) {
        int lCount = 0;
        String j = pals.get(i);
        for (int k = 0; k < pals.size(); k++) {
            if (j.equals(pals.get(k))) {
                lCount += 1;

            }
            counts[i] = lCount * pals.get(i).length();
        }

    }

    int hov = 0;
    for (int i = 0; i < pals.size(); i++) {
        if (counts[i] > hov) {
            hov = counts[i];
        }
    }
    System.out.println(hov);
}

}

Answer 1

你只需要稍微重构一下你的代码

1. 首先，你必须收集所有唯一的回文，并计算它们在字符串中出现的次数
2. 在地图上迭代，数不清回文长度和它的出现（即找到每个唯一回文的总长度）
3. 检索最大值

---

public class Palindrome {

    public static void main(String[] args) {
        try (Scanner scan = new Scanner(System.in)) {
            System.out.println(getMaxOccurrenceValue(scan.nextLine()));
        }

    }

    /** Retrieve maximum occurrence value */
    public static int getMaxOccurrenceValue(String pal) {
        return getAllPalindromes(pal).entrySet().stream()
                                     .map(entry -> entry.getKey().length() * entry.getValue())
                                     .mapToInt(Integer::intValue)
                                     .max().orElse(0);
    }

    /** Retrieve all unique palindromes for given str with occurrence amount of each palindrome */
    private static Map<String, Integer> getAllPalindromes(String str) {
        Map<String, Integer> map = new TreeMap<>();

        for (int i = 0; i < str.length(); i++) {
            for (int j = i + 1; j < str.length(); j++) {
                String sub = str.substring(i, j);

                if (isPalindrome(sub))
                    map.compute(sub, (key, count) -> Optional.ofNullable(count).orElse(0) + 1);
            }
        }

        return map;
    }

    /** Check is given str palindrome or not */
    private static boolean isPalindrome(String str) {
        for (int i = 0, j = str.length() - 1; i < j; i++, j )
            if (str.charAt(i) != str.charAt(j))
                return false;
        return true;
    }

}