187. Repeated DNA Sequences

/*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"].*//*A=0110 0001 C=0110 0011G=0110 0111T=0111 0100查找每十个字符是否重复出现，如果每次移动一个字符进行遍历时间复杂度：O(N^2)如果把每十个字符存储在hash表中，然后遍历查找时间复杂度：O(N)；由于只有四个字符，且四个字符后三个bit都不同，对四个字符重新编码，用其ASICC码的后三个bit表示，然后十个字符只需要30bit，使用一个int就可以存储，然后存储到hash表中。*/#include <iostream>#include <vector>#include <unordered_map>#include <string>#include <map>using namespace std;class Solution {public:    vector<string> findRepeatedDnaSequences(string s) {        vector<string> res;        int i=0,t=0;        int n=s.size();        cout<<n<<endl;        if(n<=10) return res;        while(i<9)            t= t<<3 | (s[i++] & 7);        unordered_map<int,int> m;        while(i<n)        {            t= (t<<3 | (s[i] & 7)) & 0x3fffffff;            m[t]++;            if(m[t]==2)            {                res.push_back(s.substr(i-9,10));            //  cout<<s.substr(i-9,10)<<endl;            }            i++;        }        return res;    }};int main(){    Solution mys;    string s="AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT";    vector<string> res=mys.findRepeatedDnaSequences(s);    for(vector<string>::iterator it=res.begin();it!=res.end();it++)        cout<<*it<<endl;    return 0;}