Java實(shí)現(xiàn)基于滑動(dòng)窗口和哈希的雙重回文檢測(cè)

在Java中，我們可以使用滑動(dòng)窗口和哈希的方法來實(shí)現(xiàn)雙重回文檢測(cè)。以下是一個(gè)簡(jiǎn)單的示例：

import java.util.HashMap;
import java.util.Map;

public class DoublePalindromeChecker {

    public static void main(String[] args) {
        String s1 = "ababa";
        String s2 = "babad";
        System.out.println("Is \"" + s1 + "\" a double palindrome? " + isDoublePalindrome(s1));
        System.out.println("Is \"" + s2 + "\" a double palindrome? " + isDoublePalindrome(s2));
    }

    public static boolean isDoublePalindrome(String s) {
        if (s == null || s.length() < 2) {
            return false;
        }

        Map<Long, Integer> prefixHash = new HashMap<>();
        long hash = 0;
        int base = 131;
        int mod = (int) 1e9 + 7;

        for (int i = 0; i < s.length(); i++) {
            hash = (hash * base + s.charAt(i)) % mod;
            prefixHash.put(hash, i + 1);
        }

        for (int i = 0; i < s.length() / 2; i++) {
            long leftHash = getHash(s, i, i);
            long rightHash = getHash(s, i, i + 1);

            if (!prefixHash.containsKey(leftHash) || !prefixHash.containsKey(rightHash) || prefixHash.get(leftHash) != i + 1 || prefixHash.get(rightHash) != i + s.length() - 1) {
                return false;
            }
        }

        return true;
    }

    private static long getHash(String s, int l, int r) {
        long leftHash = 0;
        long rightHash = 0;
        long base = 131;
        int mod = (int) 1e9 + 7;

        for (int i = l; i <= r; i++) {
            leftHash = (leftHash * base + s.charAt(i)) % mod;
            rightHash = (rightHash * base + s.charAt(r - i)) % mod;
        }

        return leftHash * mod + rightHash;
    }
}

這個(gè)程序首先計(jì)算字符串的前綴哈希值，然后使用滑動(dòng)窗口檢查每個(gè)子串是否為回文。如果所有子串都是回文，則字符串是雙重回文。

注意：這個(gè)實(shí)現(xiàn)僅適用于較短的字符串，因?yàn)楣Ｓ?jì)算可能會(huì)導(dǎo)致沖突。對(duì)于較長(zhǎng)的字符串，可以考慮使用更高效的哈希算法，如Rabin-Karp算法。