leetcode[187]Repeated DNA Sequences

標籤：

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"].

class Solution {public:/** * 所有DNA都是由一系列堿基構成, 分別為ACGT, 題目要求找出所有長度為10的子串, 這些子串在原串中出現次數必須大於1次(重複出現) * 思路： *     1、暴力枚舉肯定是會逾時 *     2、hash *        1)unordered_set<string> repeated 儲存長度為10的子字串，遍曆字串，在repeated中尋找S[i]~S[i+9]構成的子串:
 *          若未尋找到，則將其添加到repeated中，若找到，則重複，將其添加到vector<string> res中; *        2)然而unordered_set<string>對於超長的輸入串, 會消耗大量的儲存空間;  *          改進：字串壓縮（10個字元char的子串需要8bit*10=80bit，而A C G T 四個字元需要兩位bit編碼00 01 10 11，10個char字元需要2bit*10=20bit，1 int=32 bit） *        3)另外還需要考慮res中的重複答案, 因為每次只要出現在repeated中就放入res, 這顯然會造成重複放置問題; *          改進：再構造一個unordered_set<int> check, 用於儲存已經存入res中的重複子串對應的strInt值; * */    vector<string> findRepeatedDnaSequences(string s) {        vector<string> res;        if(s.empty() || s.size()<10) return res;        unordered_map<char, unsigned int> smap = {{‘A‘, 0},{‘C‘, 1},{‘G‘, 2},{‘T‘, 3}};         unordered_set<unsigned int> repeated, check;        int strInt = 0;        for(int i = 0; i < 10; i++){            strInt = (strInt<<2) + smap[s[i]];        }        repeated.insert(strInt);        for(int i = 10; i < s.size(); i++ ){            strInt = ((strInt & 0x3ffff)<<2)+smap[s[i]];            if(repeated.find(strInt)==repeated.end()){                repeated.insert(strInt);            }else{                if(check.find(strInt) == check.end()){                    res.push_back(s.substr(i-9,10));                    check.insert(strInt);                }            }        }        return res;    }};

 

leetcode[187]Repeated DNA Sequences