187. Repeated DNA Sequences(重複的DNA序列)

187. Repeated DNA Sequences(重複的DNA序列)

1 問題描述

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.

枚舉DNA字符串的中所有長度爲10的子串，將其插入到哈希Map中，並記錄子串數量;遍歷哈希map，將所有出
現超過1次的子串存儲到結果中。算法複雜度O(n)

Example:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"

Output: ["AAAAACCCCC", "CCCCCAAAAA"]

2 C++

int g_hash_map[1048576] = {0}; //2^10=1048576
class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        vector<string> result;
        if(s.length() < 10)
            return result;
        for(int i=0; i<1048576; i++)
            g_hash_map[i] = 0;

        int char_map[128] = {0};
        char_map['A'] = 0;
        char_map['C'] = 1;
        char_map['G'] = 2;
        char_map['T'] = 3;

        int key = 0;
        for(int i=9; i>=0; i--) {
            key = (key<<2) + char_map[s[i]];
        }
        g_hash_map[key] = 1;
        for(int i=10; i<s.length(); i++) {
            key = key>>2;
            key = key | (char_map[s[i]] << 18);
            g_hash_map[key]++;
        }
        for(int i=0; i<1048576; i++) {
            if(g_hash_map[i] > 1) {
                result.push_back(change_int_to_DNA(i));
            }
        }
        return result;
    }
private:
    string change_int_to_DNA(int DNA) {
        static const char DNA_CHAR[] = {'A', 'C', 'G', 'T'};
        string str;
        for(int i=0; i<10; i++) {
            str += DNA_CHAR[DNA & 3];
            DNA = DNA >> 2;
        }
        return str;
    }
};