Repeated DNA Sequences

题目描述：

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.

我自己写的代码如下，写的时候感觉就要出事，但是竟然ac了。可见leetcode并没有对内存多大限制。

public List<String> findRepeatedDnaSequences(String s) {
	List<String> result=new Stack<String>();
	int n=s.length();
	Map<String,Integer> map=new HashMap<String, Integer>();
	for(int i=0;i<=n-10;i++){
		String substr=s.substring(i,i+10);
		if(map.containsKey(substr)&&map.get(substr)==1){
			result.add(substr);
			map.put(substr, map.get(substr)+1);
		}
		else if(!map.containsKey(substr))
			map.put(substr, 1);				
	}
	return result;
}

这里的字母只有4种，A,G,C,T.为了节约内存，我们可以将他们编号成00，01，10，11.那么10个字符串需要20bits，一个int就可以搞定。

代码如下：

public List<String> findRepeatedDnaSequences(String s) {  
    List<String> result=new ArrayList<String>();
    Map<Character,Integer> map=new HashMap<Character, Integer>();
    if(s==null || s.length() < 11) return result;  
    map.put('A', 0);
    map.put('G', 1);
    map.put('C', 2);
    map.put('T', 3);
    int hash=0;
    Set<Integer> set = new HashSet<Integer>();  
    Set<Integer> unique = new HashSet<Integer>();  
    for(int i=0;i<s.length();i++){
    	char ch=s.charAt(i);
    	hash=(hash<<2)+map.get(ch);
    	if(i<9){
    		continue;
    	}else{
    		hash&=(1<<20)-1;
    		if(set.contains(hash)&&!unique.contains(hash)){
    			result.add(s.substring(i-9,i+1));
    			unique.add(hash);
    		}else{
    			set.add(hash);
    		}
    	}
    }
    return result;
} 

同样是位操作，还可以用ASCLL表的后三位来区分他们，这样连map都不需要了，更节省空间。

public class Solution {  
    public List<String> findRepeatedDnaSequences(String s) {  
        List<String> ans = new ArrayList<String>();  
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();  
        int key = 0;  
        for (int i = 0; i < s.length(); i++) {  
            key = ((key << 3) | (s.charAt(i) & 0x7)) & 0x3fffffff;  
            if (i < 9) continue;  
            if (map.get(key) == null) {  
                map.put(key, 1);  
            } else if (map.get(key) == 1) {  
                ans.add(s.substring(i - 9, i + 1));  
                map.put(key, 2);  
            }  
        }  
        return ans;  
    }  
}