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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"].
Hash Table Bit Manipulation
class Solution { public: vector<string> findRepeatedDnaSequences(string s) { unordered_map<string,int > mp; int len = s.length(),curIdx = 0; string curStr; vector<string >ret; while(curIdx + 10<=len){ curStr = s.substr(curIdx,10); if(mp.find(curStr)!=mp.end()){ ret.push_back(curStr); } else mp[curStr] = 1; curIdx ++; } return ret; } };
处理方法是 可以是将其改为 unordered_map<int ,int >，通过 4进制的转换。另外更可以通过 bitset 再次降低内存，最后需要考虑重复问题，如果用 unordered_map 可以直接标记时候已经添加到返回vector 中了， 用 bitset 可以通过 临时变量 set<string> 存储，最后生成返回的 vector。
#include <iostream> #include <string> #include <vector> #include <unordered_map> #include <bitset> #include <set> using namespace std; //class Solution { //public: // vector<string> findRepeatedDnaSequences(string s) { // unordered_map<string,int > mp; // int len = s.length(),curIdx = 0; // string curStr; // vector<string >ret; // while(curIdx + 10<=len){ // curStr = s.substr(curIdx,10); // if(mp.find(curStr)!=mp.end()){ // ret.push_back(curStr); // } // else // mp[curStr] = 1; // curIdx ++; // } // return ret; // } //}; class Solution { public: vector<string> findRepeatedDnaSequences(string s) { bitset<1048576> bst; bst.reset(); set<string > ret; int sum=0; for(int i =0;i<10;i++) sum = sum*4 + helpFun(s[i]); bst.set(sum); for( int i=10;i<s.length();i++){ sum%=262144; sum = sum*4 + helpFun(s[i]); if(bst[sum]) ret.insert(s.substr(i-9,10)); else bst.set(sum); } return vector<string>(ret.begin(),ret.end()); } int helpFun(char c) { switch(c){ case ‘A‘: return 0; case ‘C‘: return 1; case ‘G‘: return 2; case ‘T‘: return 3; } } }; int main() { string s= "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"; Solution sol; vector<string > ret = sol.findRepeatedDnaSequences(s); for(int i=0;i<ret.size();i++) cout<<ret[i]<<endl; return 0; }
[LeetCode] Repeated DNA Sequences hash map
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原文地址：http://www.cnblogs.com/Azhu/p/4339842.html
