Repeated DNA Sequences

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题目：

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"].

分析：

首先考虑将ACGT进行二进制编码

A -> 00

C -> 01

G -> 10

T -> 11

在编码的情况下，每10位字符串的组合即为一个数字，且10位的字符串有20位；一般来说int有4个字节，32位，即可以用于对应一个10位的字符串。例如

ACGTACGTAC -> 00011011000110110001

AAAAAAAAAA -> 00000000000000000000

巧妙的设计！！！

+-的优先级高于位运算！

public class Solution {    public List<String> findRepeatedDnaSequences(String s) {List<String> result = new ArrayList<String>(); int len = s.length();if (len < 10) {return result;} Map<Character, Integer> map = new HashMap<Character, Integer>();map.put('A', 0);map.put('C', 1);map.put('G', 2);map.put('T', 3); Set<Integer> temp = new HashSet<Integer>();Set<Integer> added = new HashSet<Integer>(); int hash = 0;for (int i = 0; i < len; i++) {if (i < 9) {//each ACGT fit 2 bits, so left shift 2hash = (hash << 2) + map.get(s.charAt(i)); } else {hash = (hash << 2) + map.get(s.charAt(i));//make length of hash to be 20hash = hash &  (1 << 20) - 1;  if (temp.contains(hash) && !added.contains(hash)) {result.add(s.substring(i - 9, i + 1));added.add(hash); //track added} else {temp.add(hash);}} }return result;    }}

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