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 findRepeatedDnaSequences(String s) {
	List result=new Stack();
	int n=s.length();
	Map map=new HashMap();
	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 findRepeatedDnaSequences(String s) {  
    List result=new ArrayList();
    Map map=new HashMap();
    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 set = new HashSet();  
    Set unique = new HashSet();  
    for(int i=0;i

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 findRepeatedDnaSequences(String s) {  
        List ans = new ArrayList();  
        HashMap map = new HashMap();  
        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;  
    }  
}