LeetCode() Repeated DNA Sequences 看的非常的过瘾！

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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"].

非常好的思路： 转换成位操作。

算法分析

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

20位的二进制数，至多有2^20种组合，因此hash table的大小为2^20，即1024 * 1024，将hash table设计为bool hashTable[1024 * 1024];

vector<string> findRepeatedDnaSequences(string s) {
    int  hashMap[1048576] = {0};
    vector<string> ans;
    int len = s.size(),hashNum = 0;
    if (len < 11) 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 看的非常的过瘾！

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原文地址：http://www.cnblogs.com/yanqi110/p/4996001.html