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1 class Solution { 2 private: 3 char tbl[256]; 4 public: 5 vector<string> findRepeatedDnaSequences(string s) { 6 vector<string> res; 7 8 int len = s.size(); 9 if (len < 10) { 10 return res; 11 } 12 vector<bool> exist(1<<20, false); 13 vector<bool> add(1<<20, false); 14 15 tbl[‘A‘] = 0x00; 16 tbl[‘C‘] = 0X01; 17 tbl[‘G‘] = 0x02; 18 tbl[‘T‘] = 0x03; 19 20 int mask= (1<<20) - 1; 21 int pattern = 0; 22 23 for (int i=0; i<10; i++) { 24 pattern = mask & ((pattern << 2) | tbl[s[i]]); 25 } 26 exist[pattern] = true; 27 28 for (int i=10; i<len; i++) { 29 int start = i - 10 + 1; 30 pattern = mask & ((pattern << 2) | tbl[s[i]]); 31 if (exist[pattern] && !add[pattern]) { 32 res.push_back(s.substr(start, 10)); 33 add[pattern] = true; 34 } else { 35 exist[pattern] = true; 36 } 37 } 38 return res; 39 } 40 };
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"].
将10个连续的DNA碱基序列看着是一个10位4进制的数，这样的数共有4^10=2^20个。用两个vector<bool>来分别表示，是否存在，是否已经添加到结果中即可。
LeetCode Repeated DNA Sequences
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原文地址：http://www.cnblogs.com/lailailai/p/4440180.html
