Leetcode () Repeated DNA sequences look very enjoyable!

All DNA are composed of a series of nucleotides abbreviated as a, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it's 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"].

Very good idea: convert into bit operations.

Algorithm analysis

First consider the binary encoding of the ACGT

A-00

C-01

G-10

T-11

In the case of encoding, the combination of each 10-bit string is a number, and the 10-bit string has 20 bits; Generally speaking, int has 4 bytes, 32 bits, which can be used to correspond to a 10-bit string. For example

Acgtacgtac-00011011000110110001

AAAAAAAAAA-00000000000000000000

20-bit binary number, at most 2^20 kind of combination, so the size of hash table is 2^20, that is 1024 * 1024, the hash table is designed as BOOL hashtable[1024 * 1024];

Vector<string> findrepeateddnasequences (string s) {    int  hashmap[1048576] = {0};    Vector<string> ans;    int len = S.size (), hashnum = 0;    if (Len < one) return ans;    for (int i = 0;i < 9;++i)        hashnum = hashnum << 2 | (S[i]-' A ' + 1)% 5;    for (int i = 9;i < Len;++i)        if (Hashmap[hashnum = (Hashnum << 2 | (S[i]-' A ' + 1)% 5) & 0xfffff]++ = = 1)            ans.push_back (S.substr (i-9,10));    return ans;}

　　

Leetcode () Repeated DNA sequences look very enjoyable!