187. 重复的DNA序列

链接：

187. 重复的DNA序列

题解：

class Solution {
    using ll = unsigned long long;
    ll P = 13331;   // 取131会有哈稀冲突
public:
    vector findRepeatedDnaSequences(string s) {
        int n = s.size();
        vector f(n+1, 0), p(n+1, 0); p[0] = 1;
        for(int i=0; i
            f[i+1] = f[i] * P + s[i];
            p[i+1] = p[i] * P;
        }
 
        unordered_mapint> d;
        vector ret;
        for(int i=9; i// 枚举右端点
            int h = f[i+1] - f[i-9] * p[10];
            if(++d[h] == 2) // 第二次出现
                ret.push_back(s.substr(i-9, 10));
        }
        return ret;
    }
};
 

class Solution {
    const int L = 10;
public:
    vector findRepeatedDnaSequences(string s) {
        vector ans;
        unordered_mapint> cnt;
        int n = s.length();
        for (int i = 0; i <= n - L; ++i) {
            string sub = s.substr(i, L);
            if (++cnt[sub] == 2) {
                ans.push_back(sub);
            }
        }
        return ans;
    }
};

下面这两种通过不了：

class Solution {
public:
    vector findRepeatedDnaSequences(string s) {
        if (s.size() < 10) {
            return {};
        }
        long ten9 = 1e9 + 7;
        long hhash = 1;
        long B = 256;
        //std::unordered_map table{{'A', 0}, {'C', 1}, {'G', 2}, {'T', 3}};
        for (int i = 0; i < 9; ++i) {
            hhash = (hhash * B + s[i]) % ten9;
            //hhash = (hhash * B + table[s[i]]) % ten9;
        }
        long P = 1;
        for (int i = 0; i < 9; ++i) {
            P = P * B % ten9;
        }
        std::unordered_map<long, int> hash2index; // hash值对应的最开始字符
        //cout << hhash << endl;
        //hash2index[hhash] = 0;
        std::vector result;
        for (int i = 9; i < s.size(); ++i) {
            hhash = (hhash * B + s[i]) % ten9;
            //long prev = (hhash - table[s[i-10]] * P) % ten9 * B;
            //hhash = (prev + table[s[i]] * B % ten9) % ten9;
            //cout << "i = " << i << " prev = " << prev << " new = " << hhash << endl;
            cout << "i = " << i << " hhash = " << hhash << endl;
            if (hash2index.find(hhash) != hash2index.end()) {
                result.push_back(s.substr(i-9, 10));
            } else {
                hash2index[hhash] = i - 9;
            }
            hhash = (hhash - s[i-9] * P % ten9 + ten9) % ten9;
        }
        return result;
    }
};

class Solution {
public:
    vector findRepeatedDnaSequences(string s) {
        if (s.size() < 10) {
            return {};
        }
        long ten9 = 1e9 + 7;
        long hhash = 1;
        long B = 4;
        std::unordered_map<char, long> table{{'A', 0}, {'C', 1}, {'G', 2}, {'T', 3}};
        for (int i = 0; i < 10; ++i) {
            //hhash = (hhash * B + s[i]) % ten9;
            hhash = (hhash * B + table[s[i]]) % ten9;
        }
        int P = 1;
        for (int i = 0; i < 9; ++i) {
            P = P * B % ten9;
        }
        std::unordered_map<long, int> hash2index; // hash值对应的最开始字符
        cout << hhash << endl;
        hash2index[hhash] = 0;
        std::vector result;
        for (int i = 10; i < s.size(); ++i) {
            long prev = (hhash - table[s[i-10]] * P) % ten9 * B;
            hhash = (prev + table[s[i]] * B % ten9) % ten9;
            cout << "i = " << i << " prev = " << prev << " new = " << hhash << endl;
            if (hash2index.find(hhash) != hash2index.end()) {
                result.push_back(s.substr(i-9, 10));
            } else {
                hash2index[hhash] = i - 9;
            }
        }
        return result;
    }
};