LeetCode 1044 最长重复子串_letcode 最长相同子串-优快云博客

本文介绍了一种使用随机化和双重哈希技巧寻找字符串中最长重复子串的方法，并提供了详细的Python实现代码。通过调整参数随机生成不同的哈希值，有效避免了哈希冲突。

不得不说，今日一题，属实太难了，我看了半天题解，总算是看明白了点，直接上代码

之后还得继续复习TODO

import random


class Solution:
    def longestDupSubstring(self, s: str) -> str:
        a1, a2 = random.randint(26, 100), random.randint(26, 100)
        mod1, mod2 = random.randint(10 ** 9 + 7, 2 ** 31 - 1), random.randint(10 ** 9 + 7, 2 ** 31 - 1)
        n = len(s)
        # 编码方式，将小写字母转换成数字
        arr = [ord(c) - ord('a') for c in s]
        l, r = 1, n - 1
        length, start = 0, -1
        while l <= r:
            m = l + (r - l + 1) // 2
            idx = self.check(arr, m, a1, a2, mod1, mod2)
            # 先从左半边开始进行编码，然后再对右边进行编码
            if idx != -1:
                l = m + 1
                # 记录子字符串长度和下标
                length = m
                start = idx
            else:
                r = m - 1
        return s[start:start + length] if start != -1 else ""


    def check(self, arr, m, a1, a2, mod1, mod2):
        n = len(arr)
        aL1, aL2 = pow(a1, m, mod1), pow(a2, m, mod2)
        h1, h2 = 0, 0
        # 双重哈希
        for i in range(m):
            h1 = (h1 * a1 + arr[i]) % mod1
            h2 = (h2 * a2 + arr[i]) % mod2
        # 编码组合
        seen = {(h1, h2)}
        for start in range(1, n - m + 1):
            h1 = (h1 * a1 - arr[start - 1] * aL1 + arr[start + m - 1]) % mod1
            h2 = (h2 * a2 - arr[start - 1] * aL2 + arr[start + m - 1]) % mod2

            if (h1, h2) in seen:
                return start
            seen.add((h1, h2))
        return -1


# if __name__ == '__main__':
#     print((1, 2) == (1, 2))