[Leetcode] 187. Repeated DNA sequences for repetitive DNA sequences

All DNA are composed of a series of nucleotides abbreviated as a, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it's sometimes useful to identify repeated sequences within the DNA.

Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.

Example:

input:s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT" Output: ["AAAAACCCCC", "CCCCCAAAAA"]

All DNA is composed of a series of nucleotides, abbreviated as a, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is helpful to identify sub-sequences in DNA. Write a function to find the 10-letter length of the subsequence that has occurred multiple times.

Solution 1:hash Table + hash Set

Solution 2:hash Set

Solution 3:hash table + bit Manipulte

Java:

Public list<string> findrepeateddnasequences (String s) {        set<string> result = new HashSet ();        if (S ==null | | s.length () <2)            return new ArrayList ();        set<string> temp = new HashSet ();        for (int i=0; I<s.length ()-9; i++) {            String x = s.substring (i,i+10);            if (Temp.contains (x)) {                result.add (x);            } else                temp.add (x);                      }        return new ArrayList (result);           

Java:

Public list<string> findrepeateddnasequences (String s) {    Set seen = new HashSet (), repeated = new HashSet (); 
   for (int i = 0; i + 9 < s.length (); i++) {        String ten = s.substring (i, i +);        if (!seen.add (Ten))            Repeated.add (ten);    }    return new ArrayList (repeated);}

Java:hashmap + bits Manipulation

Public list<string> findrepeateddnasequences (String s) {    set<integer> words = new hashset<> ();    set<integer> doublewords = new hashset<> ();    List<string> RV = new arraylist<> ();    char[] map = new CHAR[26];    Map[' A '-' a '] = 0;    map[' C '-' A '] = 1;    map[' G '-' A '] = 2;    map[' T '-' A '] = 3;    for (int i = 0; i < s.length ()-9; i++) {        int v = 0;        for (int j = i; j < i + ten; j + +) {            v <<= 2;            V |= Map[s.charat (j)-' A '];        }        if (!words.add (v) && Doublewords.add (v)) {            Rv.add (s.substring (i, i +));}    }    return RV;}

Python:

Class solution (Object):    def findrepeateddnasequences (self, s): "" "        : Type s:str        : rtype:list[str] "" "        dict, Rolling_hash, res = {}, 0, [] for        i in Xrange (Len (s)):            Rolling_hash = ((Rolling_hash << 3 ) & 0X3FFFFFFF) | (Ord (S[i]) & 7)            If Rolling_hash not in Dict:                dict[rolling_hash] = True            elif Dict[rolling_hash]:                res.append (s[i-9: i + 1]) C12/>dict[rolling_hash] = False        return res

Python:

def findRepeatedDnaSequences2 (self, s): "" "        : Type s:str        : rtype:list[str]" "        l, r = [], []<  C19/>if Len (s) < 10:return []        for I in range (len (s)-9):            l.extend ([S[i:i + ten])        return [k for K, V in Collections. Counter (L). Items () if v > 1]

C++:

Class Solution {public:    vector<string> findrepeateddnasequences (string s) {        unordered_set<int> seen;        Unordered_set<int> DUP;        vector<string> result;        Vector<char> m (+);        M[' A '-' a '] = 0;        m[' C '-' A '] = 1;        m[' G '-' A '] = 2;        m[' T '-' A '] = 3;                for (int i = 0; i + ten <= s.size (); ++i) {            string substr = S.substr (i, ten);            int v = 0;            for (int j = i; j < i + ten; ++j) {//20 bits < A-bit int                v <<= 2;                V |= m[s[j]-' A '];            }            if (Seen.count (v) = = 0) {//not seen                Seen.insert (v);            } else if (Dup.count (v) = = 0) {//seen but not DUP                DUP . Insert (v);                Result.push_back (substr);            } DUP        }        return result;}    ;

　　

[Leetcode] 187. Repeated DNA sequences for repetitive DNA sequences