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Dzhoker1
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Nov 21, 2018
| # Пусть дана строка S длиной N, состоящая только из маленьких латинских букв. | ||
| # Требуется найти количество различных подстрок в этой строке. | ||
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| #Надеюсь,что я правильно понял задачу... |
| for i in range(len(s)): | ||
| for j in range(i + 1, len(s)+1): | ||
| sub_s = hash(s[i:j]) | ||
| if sub_s != hash(s): |
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Данная проверка нацелена на исключение попадания хэша всей строки целиком.
Но представьте сколько раз функция хэширования должна вычислять хэш всей строки. Не проще ли посчитать его один раз в начале?
А ещё лучше вообще его не вычислять и не делать огромное число сравнений, а просто вернуть
return len(sub) - 1
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