[LeetCode] 187. Repeated DNA Sequences 求重复的DNA序列

Posted 2021-01-11 轻风舞动

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is 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"]

所有的DNA都是由一系列核苷酸组成的，简写为 A, C, G, and T，比如："ACGAATTCCG"。当研究DNA时，识别DNA里的子序列是很有帮助的。写一个函数找出10个字母长度的出现过多次的子序列。

解法1：hash table + hash set

解法2: hash set

解法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 + 10);
        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 + 10; j++) {
            v <<= 2;
            v |= map[s.charAt(j) - ‘A‘];
        }
        if(!words.add(v) && doubleWords.add(v)) {
            rv.add(s.substring(i, i + 10));
        }
    }
    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])
                dict[rolling_hash] = False
        return res

Python:

def findRepeatedDnaSequences2(self, s):
        """
        :type s: str
        :rtype: List[str]
        """
        l, r = [], []
        if len(s) < 10: return []
        for i in range(len(s) - 9):
            l.extend([s[i:i + 10]])
        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(26);
        m[‘A‘ - ‘A‘] = 0;
        m[‘C‘ - ‘A‘] = 1;
        m[‘G‘ - ‘A‘] = 2;
        m[‘T‘ - ‘A‘] = 3;
        
        for (int i = 0; i + 10 <= s.size(); ++i) {
            string substr = s.substr(i, 10);
            int v = 0;
            for (int j = i; j < i + 10; ++j) { //20 bits < 32 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;
    }
};