1044. 最长重复子串

给你一个字符串 s ，考虑其所有 重复子串 ：即，s 的连续子串，在 s 中出现 2 次或更多次。这些出现之间可能存在重叠。

返回 任意一个 具有最长长度的重复子串。如果 s 不含重复子串，那么答案为 "" 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/longest-duplicate-substring
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

import java.awt.image.Kernel;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

class Solution {

    private final long BASE1 = 127;

    private final long MOD1 = 1000000007;

    private final long BASE2 = 131;

    private final long MOD2 = 1000000009;

    public String longestDupSubstring(String s) {
        int left = 1, right = s.length() - 1;
        String ans = "";
        while (left <= right) {
            int mid = (left + right) >> 1;
            String repeat = hasRepeat(s, mid);
            if (repeat != null) {
                ans = repeat;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }

    private long maxBase(long base, long mod, int length) {
        long ans = 1;
        for (int i = 1; i < length; ++i) {
            ans = (ans * base) % mod;
        }
        return ans;
    }

    public String hasRepeat(String s, int length) {
        Set<Long> hash1 = new HashSet<>();
        Set<Long> hash2 = new HashSet<>();
        long num1 = 0;
        long num2 = 0;
        long maxBase1 = maxBase(BASE1, MOD1, length);
        long maxBase2 = maxBase(BASE2, MOD2, length);
        for (int i = 0; i < length; ++i) {
            num1 = (num1 * BASE1 + s.charAt(i)) % MOD1;
            num2 = (num2 * BASE2 + s.charAt(i)) % MOD2;
        }
        hash1.add(num1);
        hash2.add(num2);
        for (int i = length; i < s.length(); ++i) {
            num1 = ((num1 - maxBase1 * s.charAt(i - length) % MOD1 + MOD1) * BASE1 + s.charAt(i)) % MOD1;
            num2 = ((num2 - maxBase2 * s.charAt(i - length) % MOD2 + MOD2) * BASE2 + s.charAt(i)) % MOD2;
            if (hash1.contains(num1) && hash2.contains(num2)) {
                return s.substring(i - length + 1, i + 1);
            }
            hash1.add(num1);
            hash2.add(num2);
        }
        return null;
    }
}