您的位置:首页 > 产品设计 > UI/UE

[leedcode 187] Repeated DNA Sequences

2015-08-02 23:33 253 查看
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"].


public class Solution {
public List<String> findRepeatedDnaSequences(String s) {
//因为只有4个字母,所以可以创建自己的hashkey, 每两个BITS, 对应一个 incoming character. 超过20bit 即10个字符时, 只保留20bits.
Map<Character,Integer> map=new HashMap<Character,Integer>();
map.put('A',0);
map.put('C',1);
map.put('G',2);
map.put('T',3);
List<String> res=new ArrayList<String>();
int hash=0;
Set<Integer> set=new HashSet<Integer>();
for(int i=0;i<s.length();i++){
char c=s.charAt(i);
if(i<9){
hash=(hash<<2)+map.get(c);
}else{
hash=(hash<<2)+map.get(c);
hash&=(1<<20)-1;
if(set.contains(hash)){
if(!res.contains(s.substring(i-9,i+1)))
res.add(s.substring(i-9,i+1));
}else{
set.add(hash);
}
}
}
return res;
}
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: