Leetcode: Repeated DNA Sequence
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"].
Naive 方法就是两层循环,外层for(int i=0; i<=s.length()-10; i++), 内层for(int j=i+1; j<=s.length()-10; j++), 比较两个字符串s.substring(i, i+10)和s.substring(j, j+10)是否equal, 是的话,加入到result里,这样两层循环再加equals()时间复杂度应该到O(N^3)了,TLE了
方法2:进一步的方法是用HashSet, 每次取长度为10的字符串,O(N)时间遍历数组,重复就加入result,但这样需要O(N)的space, 准确说来O(N*10bytes), java而言一个char是2 bytes,所以O(N*20bytes)。String一大就MLE
最优解:是在方法2基础上用bit operation,大概思想是把字符串映射为整数,对整数进行移位以及位与操作,以获取相应的子字符串。众所周知,位操作耗时较少,所以这种方法能节省运算时间。
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
每次向右移动1位字符,相当于字符串对应的int值左移2位,再将其最低2位置为新的字符的编码值,最后将高2位置0。
Cost分析:
时间复杂度O(N), 而且众所周知,位操作耗时较少,所以这种方法能节省运算时间。
省空间,原来10个char要10 Byte,现在10个char总共20bit,总共O(N*20bits)
空间复杂度:20位的二进制数,至多有2^20种组合,因此HashSet的大小为2^20,即1024 * 1024,O(1)
1 public class Solution { 2 public List<String> findRepeatedDnaSequences(String s) { 3 ArrayList<String> res = new ArrayList<String>(); 4 if (s==null || s.length()<=10) return res; 5 HashMap<Character, Integer> dict = new HashMap<Character, Integer>(); 6 dict.put('A', 0); 7 dict.put('C', 1); 8 dict.put('G', 2); 9 dict.put('T', 3); 10 HashSet<Integer> set = new HashSet<Integer>(); 11 HashSet<String> result = new HashSet<String>(); //directly use arraylist to store result may not avoid duplicates, so use hashset to preselect 12 int hashcode = 0; 13 for (int i=0; i<s.length(); i++) { 14 if (i < 9) { 15 hashcode = (hashcode<<2) + dict.get(s.charAt(i)); 16 } 17 else { 18 hashcode = (hashcode<<2) + dict.get(s.charAt(i)); 19 hashcode &= (1<<20) - 1; 20 if (!set.contains(hashcode)) { 21 set.add(hashcode); 22 } 23 else { 24 //duplicate hashcode, decode the hashcode, and add the string to result 25 String temp = s.substring(i-9, i+1); 26 result.add(temp); 27 } 28 } 29 } 30 for (String item : result) { 31 res.add(item); 32 } 33 return res; 34 } 35 }
naive方法:
1 public class Solution { 2 public List<String> findRepeatedDnaSequences(String s) { 3 ArrayList<String> res = new ArrayList<String>(); 4 if (s==null || s.length()<=10) return res; 5 for (int i=0; i<=s.length()-10; i++) { 6 String cur = s.substring(i, i+10); 7 for (int j=i+1; j<=s.length()-10; j++) { 8 String comp = s.substring(j, j+10); 9 if (cur.equals(comp)) { 10 res.add(cur); 11 break; 12 } 13 } 14 } 15 return res; 16 } 17 }