leetcode No187. Repeated DNA Sequences
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Question:
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"].
查找重复的子串,子串长度为10
Algorithm:
字符串中只包含4个字符:A, C, G, T。它们的ASCII码的二进制形式是:
- A : 0100 0001
- C : 0100 0011
- G : 0100 0111
- T : 0101 0100
这4个字符的末3位是不同的,因此,我们可以 3个bits 来表示其中的一个字符。
又因为每个子串的长度为10,因此总的位数是:10 x 3 = 30,一个int就足够存放它。
Accepted Code:
class Solution {public:/* A:00 ->0 C:01 ->1 G:10 ->2 T:11 ->3*/ int encode(string str) // { int res=0; for(int i=0;i<str.size();i++) { res=res<<3|(str[i]&0x07); } return res; } vector<string> findRepeatedDnaSequences(string s) { unordered_map<int,int> hash; vector<string> res; if(s.size()<10) return res; for(int i=0;(i+9)<s.size();i++) { if(hash[encode(s.substr(i,10))]==1) { res.push_back(s.substr(i,10)); hash[encode(s.substr(i,10))]++; } else hash[encode(s.substr(i,10))]++; } return res; }};
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