187. Repeated DNA Sequences (String; Bit)

时间:2022-08-21 17:29:11

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"].

思路I:遍历string,每次截取10个字符,判断出现次数。

Result: Time Limit Exceeded

思路II:字符数较少=>用数字表示字符=>用bitmap来表示字符串,好处:节省空间

比如本题只可能出现4种字符=>可表示为0,1,2,3,即可以用2bits来表示=>字符原本一个字符占1 byte = 8 bits,现在只要2 bits

class Solution {
public:
int getVal(char ch) {
if (ch == 'A') return ;
if (ch == 'C') return ;
if (ch == 'G') return ;
if (ch == 'T') return ;
} vector<string> findRepeatedDnaSequences(string s) {
int sLen = s.length();
unsigned int val=;
char mp[*]={};
vector<string> ret;
string str; if(sLen < ) return ret; for(int i = ; i < ; i++){
val <<=;
val |= getVal(s[i]);
} for(int i = ; i < sLen; i++){
val <<= ;
val |= getVal(s[i]);
val &= 0xFFFFF;
if(++mp[val] == ){
str = s.substr(i-,);
ret.push_back(str);
}
} return ret;
}
};