Java for LeetCode 187 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 = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT",

Return:
["AAAAACCCCC", "CCCCCAAAAA"].
解题思路一：

直接用HashMap实现，JAVA实现如下：

    static public List<String> findRepeatedDnaSequences(String s) {
    	List<String> list=new ArrayList<String>();
        HashMap<String,Integer> hm=new HashMap<String,Integer>();
        for(int i=0;i<=s.length()-10;i++){
        	if(hm.containsKey(s.substring(i,i+10)))
        		list.add(s.substring(i,i+10));
        	else hm.put(s.substring(i,i+10), 1);
        }
        return list;
    }

结果Memory Limit Exceeded

解题思路二：

模拟Hash，将A、C、G、T分别变为0、1、2、3，然后每10位计算下hashcode，如果hashcode所在的count为1则输出，JAVA实现如下：

	static int getValue(char ch) {
		if (ch == ‘A‘)
			return 0;
		else if (ch == ‘C‘)
			return 1;
		else if (ch == ‘G‘)
			return 2;
		else
			return 3;
	}
	static public List<String> findRepeatedDnaSequences(String s) {
		List<String> list = new ArrayList<String>();
		if (s.length() <= 10)
			return list;
		int[] count = new int[(1 << 20)-1];
		int hash = 0;
		for (int i = 0; i < 9; i++)
			hash = (hash << 2) | getValue(s.charAt(i));
		for (int i = 9; i < s.length(); i++) {
			hash = (1<<20)-1&((hash << 2) | getValue(s.charAt(i)));
			if (count[hash]==1)
				list.add(s.substring(i - 9, i + 1));
			count[hash]++;
		}
		return list;
	}