187. Repeated DNA Sequences Medium

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.

Given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT",Return:["AAAAACCCCC", "CCCCCAAAAA"].

思路:我们需要寻找的是重复出现过的子串，看到题目下方的提示tags 说使用哈希表和位运算，因此我们想到使用数字来表示A,C,G,T,分别将他们标为0，1，2，3，我们可以用总共20位二进制数来表示长度为十的子串，并且用一个int来表示这个二进制数，这样只要两个子串有相同的int，我们就将其放在vector中作为结果输出，由于题目要求输出的子串中没有重复，所以我们使用map来排除重复的情况。

class Solution {public:    vector<string> findRepeatedDnaSequences(string s) {        map<string, string>result;        if (s.length() <= 10) {            return vector<string>();        }        int tran[s.length()];        for (int i = 0; i < s.length(); i++) {            if (s[i] == 'A') {                tran[i] = 0;            }            if (s[i] == 'C') {                tran[i] = 1;            }            if (s[i] == 'G') {                tran[i] = 3;            }            if (s[i] == 'T') {                tran[i] = 2;            }        }        map<int,int>hash;        int sum = 0;        int i = 0;        while (i < 9) {          sum = (sum << 2) | (tran[i]);          i++;        }        for (; i < s.length(); i++) {            sum = ((sum & 0x3ffff) << 2) | (tran[i]);            map<int,int>::iterator it = hash.find(sum);            if (it == hash.end()) {                hash[sum] = 1;            } else {                result[s.substr(i - 9, 10)] = s.substr(i - 9, 10);            }        }        vector<string>ans;        map<string,string>::iterator it = result.begin();        while (it != result.end()) {            ans.push_back(it -> second);            it++;        }        return ans;    }};