LeetCode – Refresh – Repeated DNA Sequences

Follow the hint from https://leetcode.com/discuss/24478/i-did-it-in-10-lines-of-c

 

1. The last three digits of binary format will be :

A   001

C   011

G   111

T　100

 

So as 10 chars need 3 * 10 bits to record the status. We choose 0x3FFFFFF, still less than 32 bits.

9 chars: 0x1FFFFFFF

8 chars:    0xFFFFFF

7              0x1FFFFF

6              0x3FFFF

5              0x7FFF

4　　        0xFFF

3              0x1FF

2              0x3F

1              0x7

 

Code:

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        vector<string> result;
        unordered_map<int, int> record;
        int current = 0, len = s.size();
        for (int i = 0; i < len; i++) {
            if (record[current = (current << 3 & 0x3FFFFFFF | s[i] & 7)]++ == 1) {
                result.push_back(s.substr(i-9, 10));
            }
        }
        return result;
    }
};