注:这种方法是在leetcode的讨论区看到的,所以说是转载,如有不妥,私信删除。
题目描述:
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"].
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思路:
ACGT四个字母的ASCII值分别是65,67,71,84,暂且将四个字母的ASCII值设为x,所以(x-'A'+1)%5分别为1,3,2,0,也就是说每个字母只需要2bit就可以表示,20个bit就可以表示长度为10的字符串。
代码:
class Solution {
public:
vector<string> findRepeatedDnaSequences(string s) {
vector<string> res;
unordered_map<int,int> hash;
int hashnum=0;
int len=s.size();
if (len<11)
return res;
for (int i=0; i<9; i++)
hashnum=hashnum<<2|(s[i]-'A'+1)%5;
for (int i=9;i<len; i++)
{
hashnum=(hashnum<<2|(s[i]-'A'+1)%5)&0xfffff;
if (hash[hashnum]++==1)
res.push_back(s.substr(i-9,10));
}
return res;
}
};