LeetCode Repeated DNA Sequences

原题链接在这里：https://leetcode.com/problems/repeated-dna-sequences/

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"].

题解：

取长度为10的字符串，比较是否在之前出现过，这里可以用HashSet来保留之前出现过的substring.

也可以在把这长度为10的字符串生成hashcode来检查是否有重复.算hashcode时，A为00; C为01; G为10; T为11. 每次hash位运算左移两位，后两位就变成0了，再和新的字符代表数字做位运算或。 

Time Complexity: O(n). n = s.length().

Space: O(n).

AC Java:

class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> res = new ArrayList<String>();
        if(s == null || s.length() < 10){
            return res;
        }
        
        HashSet<Integer> resSet = new HashSet<Integer>();
        HashSet<Integer> hs = new HashSet<Integer>();
        char [] map = new char[26];
        map['A'-'A'] = 0;
        map['C'-'A'] = 1;
        map['G'-'A'] = 2;
        map['T'-'A'] = 3;
        for(int i = 0; i<=s.length()-10; i++){
            int hash = 0;
            for(int j = i; j<i+10; j++){
                hash = (hash << 2) | map[s.charAt(j)-'A'];
            }
            
            if(!hs.add(hash) && resSet.add(hash)){
                res.add(s.substring(i, i+10));
            }
        }
        
        return res;
    }
}