Repeated DNA Sequences

题目：

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：

这个题上来打算用hashmap把所有的子序列都放进去，找相同的。是不是有点天真了。。

查了一下，可以用bit mask来做，恍然大悟。bit manipulation总是可以让代码变得很简洁。

直接看代码吧。也没什么巧妙地地方，只是用两个bit代表一个字母。

A：00 

C：01

G:  10

T:  11

    public List<String> findRepeatedDnaSequences(String s) {
        List<String> res = new ArrayList<String>();
        if (s.length() <= 10) {
            return res;
        }
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        map.put('A', 0);
        map.put('C', 1);
        map.put('G', 2);
        map.put('T', 3);
        
        int mask = 0;
        
        Set<Integer> subSeq = new HashSet<Integer>();
        Set<Integer> addedSubSeq = new HashSet<Integer>();
        
        for (int i = 0; i < s.length(); i++) {
            if (i < 9) {
                mask = mask << 2;
                mask += map.get(s.charAt(i));
            }else{
                mask = mask << 2;
                mask += map.get(s.charAt(i));
                
                //we only need 20 bits for 10 letter
                mask = mask << 12;
                mask = mask >>> 12;
                if (subSeq.contains(mask) && !addedSubSeq.contains(mask)) {
                    res.add(s.substring(i - 9, i + 1));
                    addedSubSeq.add(mask);
                } else {
                    subSeq.add(mask);
                }
            }
        }
        return res;
    }