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"].
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)
follow up : 如果是inorder 的话用radix sort
follow up 如果是scanner:
Scanner scanner=new Scanner(System.in);
char a=scanner.nextCharacter();
或 String a=scanner.next();//注意不是nextString()
public static int[] RadixSort(int[] ArrayToSort, int digit)
{
//low to high digit
for (int k = 1; k <= digit; k++)
{
//temp array to store the sort result inside digit
int[] tmpArray = new int[ArrayToSort.Length];
//temp array for countingsort
int[] tmpCountingSortArray = new int[10]{0,0,0,0,0,0,0,0,0,0};
//CountingSort
for (int i = 0; i < ArrayToSort.Length; i++)
{
//split the specified digit from the element
int tmpSplitDigit = ArrayToSort[i]/(int)Math.Pow(10,k-1) - (ArrayToSort[i]/(int)Math.Pow(10,k))*10;
tmpCountingSortArray[tmpSplitDigit] += 1;
}
for (int m = 1; m < 10; m++)
{
tmpCountingSortArray[m] += tmpCountingSortArray[m - 1];
}
//output the value to result
for (int n = ArrayToSort.Length - 1; n >= 0; n--)
{
int tmpSplitDigit = ArrayToSort[n] / (int)Math.Pow(10,k - 1) - (ArrayToSort[n]/(int)Math.Pow(10,k)) * 10;
tmpArray[tmpCountingSortArray[tmpSplitDigit]-1] = ArrayToSort[n];
tmpCountingSortArray[tmpSplitDigit] -= 1;
}
//copy the digit-inside sort result to source array
for (int p = 0; p < ArrayToSort.Length; p++)
{
ArrayToSort[p] = tmpArray[p];
}
}
return ArrayToSort;
}
As our alphabet A consists of only 4 letters we can be not afraid of collisions. The hash for a current window slice could be found in a constant time by subtracting the former first character
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;
}
}