【刷题-LeetCode】187 Repeated DNA Sequences

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.

Example:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"

Output: ["AAAAACCCCC", "CCCCCAAAAA"]

解法1 hash + slice window，记录出现的每个子串，找出重复出现的子串

时间复杂度：(O(N-L))

空间复杂度：(O((N-L)L))

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        vector<string>ans;
        if(s.size() < 10)return ans;
        unordered_map<string, int>mp;
        for(int i = 0; i < s.size() - 10 + 1; ++i){
            string tmp = s.substr(i, 10);
            if(mp[tmp] == 1)ans.push_back(tmp);
            mp[tmp] += 1;
        }
        return ans;
    }
};

解法2 将字符串序列看作是数字序列，在本题中将其看做4进制数。对于每一个长度为(L)子串，都可以将其转换为一个数字，以此来优化解法1的空间复杂度

对于字符串([s_i , s_{i+1}, ..., s_{i+L-1}])：(h_i = sum_{k=i}^{i+L-1} c_{j}4^{L+i-1-j})

对于字符串([s_{i+1},s_{i+2}, ..., s_{i+L-1}, s_{i+L}])：

[h_{i+1} = sum_{j=i+1}^{i+L} c_{j}4^{L+i-j}=(h_i imes L - c_i imes 4^L) + c_{i+L} ]

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        vector<string>ans;
        if(s.size() < 10)return ans;
        unordered_map<long long int, int>mp;
        unordered_map<char, int>basis{{'A', 0}, {'G',1},{'T', 2},{'C', 3}};
        long long int h = 0, d = pow(4, 10);
        for(int i = 0; i < 10; ++i)h = h * 4 + basis[s[i]];
        mp[h] = 1;
        for(int i = 1; i < s.size() - 10 + 1; ++i){
            h = (h * 4 - basis[s[i-1]] * d) + basis[s[i+10-1]];
            if(mp[h] == 1)ans.push_back(s.substr(i, 10));
            mp[h] += 1;
        }
        return ans;
    }
};