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)

 public class Solution {
public List<String> findRepeatedDnaSequences(String s) {
ArrayList<String> res = new ArrayList<String>();
if (s==null || s.length()<=10) return res;
HashMap<Character, Integer> dict = new HashMap<Character, Integer>();
dict.put('A', 0);
dict.put('C', 1);
dict.put('G', 2);
dict.put('T', 3);
HashSet<Integer> set = new HashSet<Integer>();
HashSet<String> result = new HashSet<String>(); //directly use arraylist to store result may not avoid duplicates, so use hashset to preselect
int hashcode = 0;
for (int i=0; i<s.length(); i++) {
if (i < 9) {
hashcode = (hashcode<<2) + dict.get(s.charAt(i));
}
else {
hashcode = (hashcode<<2) + dict.get(s.charAt(i));
hashcode &= (1<<20) - 1;
if (!set.contains(hashcode)) {
set.add(hashcode);
}
else {
//duplicate hashcode, decode the hashcode, and add the string to result
String temp = s.substring(i-9, i+1);
result.add(temp);
}
}
}
for (String item : result) {
res.add(item);
}
return res;
}
}

naive方法:

 public class Solution {
public List<String> findRepeatedDnaSequences(String s) {
ArrayList<String> res = new ArrayList<String>();
if (s==null || s.length()<=10) return res;
for (int i=0; i<=s.length()-10; i++) {
String cur = s.substring(i, i+10);
for (int j=i+1; j<=s.length()-10; j++) {
String comp = s.substring(j, j+10);
if (cur.equals(comp)) {
res.add(cur);
break;
}
}
}
return res;
}
}
05-11 20:02
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