leetcode:Repeated DNA Sequences

本文介绍了一个函数,用于在DNA分子中查找所有长度为10的重复子串序列。

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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.

For example,

Given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT",

Return:
["AAAAACCCCC", "CCCCCAAAAA"].

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class Solution {

private:
    int char2val(char c) {
        switch (c) {
            case 'A': return 0;
            case 'C': return 1;
            case 'G': return 2;
            case 'T': return 3;
        }
    }

public:
    vector<string> findRepeatedDnaSequences(string s) {
     
        vector<string> ans;
        
        if (s.size() <= 10)
            return ans;
            
        //map<int, int> hashTable;
        char  hashTable[1048576] = {0};
        int len = s.size();
        int hashValue = 0;
        int mask = (1<<20)-1;
            
        for (int i=0; i<9; i++)
            hashValue = (hashValue << 2) | char2val(s[i]);
            
        for (int i=9; i<len; i++)
        {
            hashValue = (hashValue << 2 | char2val(s[i])) & mask;
            if (hashTable[hashValue]++ == 1)
            {
                ans.push_back(s.substr(i-9, 10));
            }
        }
        
        return ans;
        
        /*
        char  hashMap[1048576] = {0};
        vector<string> ans;
        int len = s.size(),hashNum = 0;
        if (len < 11) return ans;
        for (int i = 0;i < 9;++i)
            hashNum = hashNum << 2 | (s[i] - 'A' + 1) % 5;
        for (int i = 9;i < len;++i)
            if (hashMap[hashNum = (hashNum << 2 | (s[i] - 'A' + 1) % 5) & 0xfffff]++ == 1)
                ans.push_back(s.substr(i-9,10));
        return ans;  
        */
    }
};


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