[LeetCode] 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”].

• 解题思路
  按照提示，采用Hash Table和Bit Manipulation来处理。
  备注：采用hash_map来做时，自己电脑上可以编译通过，但是LeetCode提示不存在hash_map对象。故改为使用map，但效率较低。
• 实现代码

/************************************************************* * @Author : 楚兴 * @Date : 2015/2/8 11:07 * @Status : Accepted * @Runtime : 238 ms *************************************************************/
#include <vector>
#include <iostream>
#include <string>
#include <map>
using namespace std;

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        vector<string> result;
        if (s.length() <= 10)
        {
            return result;
        }
        map<int,int> mymap;
        int i = 0;
        int cursor = 0;
        while (i < 9)
        {
            cursor = cursor << 3 | s.at(i) & 7;  //优先级顺序<<、&、|
            i++;
        }

        int mask = 0x7FFFFFF;
        while (i < s.length())
        {
            //cursor & mask得到27bit
            cursor = (cursor & mask) << 3 | s.at(i) & 7;  
            i++;
            auto it = mymap.find(cursor);
            if (it != mymap.end())
            {
                int count = (*it).second;
                if (count == 1)
                {
                    result.push_back(s.substr(i - 10, 10));
                }

                get<1>(*it) = count + 1; //更改second的值
            }
            else
            {
                mymap.insert(make_pair(cursor,1));
            }
        }

        return result;
    }
};

int main()
{
    Solution s;
    string str = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT";
    vector<string> result = s.findRepeatedDnaSequences(str);
    for (auto it = result.begin(); it != result.end(); it++)
    {
        cout<<(*it).c_str()<<endl;
    }
    system("pause");
}