Created
February 7, 2021 06:48
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from typing import List | |
class Solution: | |
def findRepeatedDnaSequences(self, s: str) -> List[str]: | |
import collections | |
table = {"A":0, "C":1, "G":2, "T": 3} # 2bit表現 | |
if s is None: | |
return [] | |
ll = len(s) | |
if ll < 10: | |
return [] | |
hashVal = 0 # 今までのハッシュ値 | |
hashHistroy = collections.defaultdict(int) # unordered_map<int,int> | |
hashStr = dict() # 結果表示用のunordered_map<int,str> | |
# 最初の10文字を計算する | |
for i in range(10): | |
hashVal <<= 2 | |
hashVal |= table[s[i]] | |
hashVal &= ( (1<<20) - 1) # bitmask 011111111... | |
# 最初の10文字の処理 | |
hashHistroy[hashVal] += 1 | |
hashStr[hashVal] = s[0:10] | |
for i in range(1, ll - 9): # 1文字目から最後までループ | |
hashVal <<= 2 # 抜ける数字をシフトして | |
hashVal &= ( (1<<20) - 1) # 20bitだけ残す = 抜ける数字を消す | |
hashVal |= table[s[i+9]] # 新しく入る数字をORで入れる | |
hashHistroy[hashVal] += 1 # このハッシュの回数をカウント | |
hashStr[hashVal] = s[i:i + 10] # 今の10文字が答えになったときのために保存 | |
# 復元。ハッシュのカウントを数えて、2回以上なら、結果として表示 | |
res = [] | |
for k in hashHistroy.keys(): | |
if hashHistroy[k] > 1: | |
res.append(hashStr[k]) | |
return res |
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