Leetcode 187. Repeated DNA Sequences[medium]

题目：
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”].

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先吐槽一下，在leetcode上做题真的是很烦，在内存上各种做文章，感觉真的很没意思。
吐槽完毕。

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这个题因为一共只有4中字母，所以可以用2位二进制位表示，所以10个字母总共20个二进制位，可以直接用数组存储，判重。
本来一个int型vis数组就可以搞定，可是leetcode显示MLE。所以就加了一个ck数组辅助vis进行记录。

class Solution {public:    bool vis[(1 << 20) - 1];    bool ck[(1 << 20) - 1];    vector<string> findRepeatedDnaSequences(string s) {        map<char, int> mp;        mp['A'] = 0;        mp['T'] = 1;        mp['C'] = 2;        mp['G'] = 3;        memset(vis, 0, sizeof vis);        memset(ck, 0, sizeof ck);        int mask = (1 << 20) - 1;        vector<string> ans;        int bc = 0;        if (s.length() < 10) return ans;        for (int i = 0; i < 10; i++) {            bc = (bc << 2) | mp[s[i]];        }        vis[bc]++;        for (int i = 10; i < s.length(); i++) {            bc = ((bc << 2) & mask) | mp[s[i]];            if (vis[bc]) {                if (!ck[bc]) {                    tostr(bc, ans);                    ck[bc] = true;                }            }            else vis[bc] = true;        }         return ans;    }    void tostr(int x, vector<string>& ans) {        string str;        int y;        for (int i = 0; i < 10; i++) {            y = x & 3;            x >>= 2;            if (y == 0) str.push_back('A');            else if (y == 1) str.push_back('T');            else if (y == 2) str.push_back('C');            else str.push_back('G');        }        for (int i = 0; i < str.length() / 2; i++) {            swap(str[i], str[str.length() - i - 1]);        }        ans.push_back(str);    }};