比较两个字符串中连续几个字符相同的

jihuyao

Happened to read article about bGPT/Transformer in sins.com today.  Briefly, it is true no need to check byte one by one if, say bind 4 bytes in sequence and index it by a larger number, chr(1)*1000000+chr(2)*10000+chr(3)*100+chr(4).  So one may compare 4 bytes every time but remember to extract other strings in between use math and check instr in s2.  Any work done upfront will save time later.  In fact, the whole string s1 or s2 can be considered as a large number (if resource allows), and loop processes are a bunch of math operations.  After all, forget coding and just ask chatGTP to do the dirty work to same time and money.  lol

newkid

jihuyao 发表于 2024-3-13 09:54
Happened to read article about bGPT/Transformer in sins.com today.  Briefly, it is true no need to c ...

你是想一下子比较多个字符，那么你就要采用一种N进制表示法，N要足够大，还要考虑字符有重复的情况如何处理。还需要把所有子串都进行预处理。但是这个问题中，子串的长度是未知的，这个技巧也用不上。