[Leetcode] 187 Repeated DNA Sequences, leetcoderepeated

[Leetcode] 187 Repeated DNA Sequences, leetcoderepeated

(1) The first practice is to use map <string, int> to record the number of strings with 10 characters. If it exceeds 2, push_back is entered into ans. But MLE indicates that using string is not a good method.

Below is the MLE code:

class Solution {public:    vector<string> findRepeatedDnaSequences(string s) {    vector <string>  ans;        map<string,int> mp;        if(s.length()<10)        return ans;        for(int i=0;i<s.length()-10;i++)        mp[s.substr(i,10)]++;        map<string,int>::iterator it;        for(it=mp.begin();it!=mp.end();++it)        {        if(it->second>1)        ans.push_back(it->first);        }        return ans;    }};

(2) After reading the Tags, I am prompted to use the bitwise operation. This reminds me of the uniqueness of the prefix code of the Hoffmann code, so the following mark can be used here:

A: 00 T: 01 C: 10G: 11 A total of 10 characters, A total of 20 characters, and an int has 32 characters, so map <int, int> processing can reduce space usage.

The clock maintains a 20-bit space. During traversal, the first character is removed from the left 14-bit space, and the first character is removed from the right 12-bit space, in this way, it has played a role in saving time.

class Solution {public:    vector<string> findRepeatedDnaSequences(string s) {    vector <string>  ans;        map <int,int> mp;        map <char,int> cur;        set<string> st;        cur['A']=0;        cur['T']=1;        cur['C']=2;        cur['G']=3;        if(s.length()<10)        return ans;        int temp;        for(int i=0;i<9;i++)        {        temp<<=2;        temp|=cur[s[i]];         }       // mp[temp]++;        for(int i=9;i<s.length();i++)        {        temp<<=14;        temp>>=12;        temp|=cur[s[i]];        mp[temp]++;        if(mp[temp]>=2)        st.insert(s.substr(i-9,10));        }        set<string>::iterator it;        for(it=st.begin();it!=st.end();it++)        ans.push_back(*it);                return ans;    }};

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