Repeated DNA Sequences

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 = "AAAAACCCCCAAAAACCCCCAAAAAGGGTTT",

Return:
["AAAAACCCCC", "CCCCCAAAAA"].


这题其实挺简单的：
1.[AAAAAAAAAAAA] 也有[AAAAAAAAAA]作为return。 substrings之间可以overlap；
2.其实就是怎么对一个length＝10的string找一个hashcode。（如果hashmap 以string 为key， 则out of memory）
再就是 如果每次去算hashcode的时候 用substring 算 也会ofm。。。。 

public class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> result = new ArrayList<String>();
        if(s == null || s.length() < 10) return result;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        Integer val = 0;
        for(int i = 0; i < 10; i ++){
            val = val << 2;
            val |= toInt(s.charAt(i));
        }
        map.put(val, 1);
        for(int i = 10; i < s.length(); i ++){
            val = ((val & 0x3ffff) << 2) | toInt(s.charAt(i));
            if(map.containsKey(val)) map.put(val, map.get(val) + 1);
            else map.put(val, 1);
        }
        for(Integer v : map.keySet())
            if(map.get(v) > 1) result.add(toDNA(v));
        return result;
    }
    
    private Integer toInt(char c){
        if(c == 'A') return 0;
        else if(c == 'C') return 1;
        else if(c== 'G') return 2;
        else return 3;//T
    }
    
    private String toDNA(Integer i){
        StringBuilder sb = new StringBuilder();
        for(int j = 0; j < 10; j ++){
            int tmp = i % 4;
            i = i / 4;
            char c = 'T';
            if(tmp == 0) c = 'A';
            else if(tmp == 1) c = 'C';
            else if(tmp == 2) c ='G';
            sb.insert(0, c);
        }
        return sb.toString();
    }
}