Leetcode: Repeated DNA Sequences

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本文介绍了一种使用哈希表和四进制映射的方法来高效查找DNA分子中长度为10的重复序列。通过将DNA序列转换为整数值并利用哈希表记录出现次数，可以在O(n)的时间复杂度内找出所有重复的10字母长序列。

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”].

因为涉及到查找重复元素的问题，所以用map做最简单。但是如果直接用string做key的话会导致内存超过限制，因此必须想点办法。考虑到只有A, C, G, T四种字符，因此可以将字符串映射成四进制整数，并用该值作为map的key值。代码如下：

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
       vector<string> result;
       int val[26] = {0};
       int max[26] = {0};
       unordered_map<int, int> mp;
       int key = 0;
       int i;

       val['C'-'A'] = 1;
       val['G'-'A'] = 2;
       val['T'-'A'] = 3;
       max['C'-'A'] = 1 << 20; 
       max['G'-'A'] = 2 << 20;
       max['T'-'A'] = 3 << 20;

       if (s.length() < 10) return result;

       for (i = 0; i < 10; ++i) {
           key <<= 2;
           key += val[s[i]-'A'];
       }
       mp[key]++;

       for (i = 10; i < s.length(); ++i) {
           key <<= 2;
           key -= max[s[i-10]-'A'] - val[s[i]-'A'];
           mp[key]++;
           if (mp[key] == 2) result.push_back(s.substr(i-9, 10));
       }

       return result;
    }
};

算法复杂度 $O(n)$