[位运算]187. Repeated DNA Sequences

题目：187. Repeated DNA Sequences［Medium]
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"].

找出DNA序列中所有长度大于10且出现次数大于1的子串。

方法一：用HashMap存储所有子串，结果：Time Limit Exceeded

class Solution {
    public List findRepeatedDnaSequences(String s) {
        List res = new ArrayList();
        if(s.length() <= 10) return res;
        
        Map strmap = new HashMap();
        int i =0;
        
        while( i <= s.length()-10){
            String temp = s.substring(i, i+10);
            if(!strmap.containsKey(temp)){
                strmap.put(temp,1);
                i++;
            }else{
                if(strmap.get(temp) == 1){
                    res.add(temp);
                    strmap.put(temp,-1) ; //had be add to res
                    i++;
                }
            }
        }
        return res;
    }
}

---

方法二：位运算
Runtime: 63 ms
对于A，C，G，T四个字符其二进制表示为如下，仅有最后三位不同。

A: 0100 0001
C: 0100 0011　　
G: 0100 0111　　
T: 0101 0100

每一位用1bit表示，10个字符供需要10x3 ＝ 30bit。一个int有32bit，可以表示一个字符串。

注意

0x7ffffff 是 111...1111 ， 一共3+6*4 = 27bit位个1
substring 的范围是前闭后开：[0, 10) 取得是－>［0，9］

class Solution {
    public List findRepeatedDnaSequences(String s) {
        List res = new ArrayList();
        if(s.length() <= 10) return res;
        
        Map strmap = new HashMap();//substring, 出现次数
        int i =0;
        int mask = 0x7ffffff; //111...1111  一共3+6*4 = 27bit位个1
        int cur =0;
        while( i< 9 ) {
            cur = cur<<3 | (s.charAt(i) & 7); i++;
        }
        //i =9
        while( i < s.length()){
            cur = ((cur & mask) << 3) | ((int)s.charAt(i) & 7); 
            //((cur & mask) << 3) ｜:取cur的后27位再补3个0，再加上i的三位
            if(!strmap.containsKey(cur)){
                strmap.put(cur,1);
            }else{
                if(strmap.get(cur) == 1){
                    res.add(s.substring(i-9,i+1)); //[i-9, i+1)
                    strmap.put(cur,-1) ; //had be add to res
                }
            }
             i++;
        }
        return res;
    }
}

方法三：
Runtime: 41 ms
在solution里看到的，更快更节约时间。

在set.add(obj)方法里，如果obj已在set中存在，该方法会返回false。

class Solution {
    public List findRepeatedDnaSequences(String s) {
        Set set = new HashSet();
        Set repeat = new HashSet();
        
        for(int i=0; i