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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.
原题链接：https://oj.leetcode.com/problems/repeated-dna-sequences/
直接字符串匹配；设计next数组，存字符串中每个字母在其中后续出现的位置；遍历时以next数组为起始。
简化考虑长度为4的字符串
src A C G T A C G T
next [4] [5] [6] [7] [-1] [-1] [-1] [-1]
那么匹配ACGT字符串的过程，匹配next[0]之后的3位字符即可
src A C G T A A C G T
next [4] [5] [6] [7] [5] [-1] [-1] [-1] [-1]
多个A字符后继，那么需要匹配所有后继，匹配next[0]不符合之后，还要匹配next[next[0]]
src A A A A A A
next [1] [2] [3] [4] [5] [-1]
重复的情况，在next[0]匹配成功时，可以把next[next[0]]置为-1，即以next[0]开始的长度为4的字符串已经成功匹配过了，无需再次匹配了；当然这么做只能减少重复的情况，并不能消除重复，因此仍需要使用一个set存储匹配成功的结果，方便去重
构造next数组的复杂度O(n^2)，遍历的复杂度O(n^2)；总时间复杂度O(n^2)
1 #include <string> 2 #include <vector> 3 #include <set> 4 5 class Solution { 6 public: 7 std::vector<std::string> findRepeatedDnaSequences(std::string s); 8 9 ~Solution(); 10 11 private: 12 std::size_t* next; 13 }; 14 15 std::vector<std::string> Solution::findRepeatedDnaSequences(std::string s) { 16 std::vector<std::string> rel; 17 18 if (s.length() <= 10) { 19 return rel; 20 } 21 22 next = new std::size_t[s.length()]; 23 24 // cal next array 25 for (int pos = 0; pos < s.length(); ++pos) { 26 next[pos] = s.find_first_of(s[pos], pos + 1); 27 } 28 29 std::set<std::string> tmpRel; 30 31 for (int pos = 0; pos < s.length(); ++pos) { 32 std::size_t nextPos = next[pos]; 33 while (nextPos != std::string::npos) { 34 int ic = pos; 35 int in = nextPos; 36 int count = 0; 37 while (in != s.length() && count < 9 && s[++ic] == s[++in]) { 38 ++count; 39 } 40 if (count == 9) { 41 tmpRel.insert(s.substr(pos, 10)); 42 next[nextPos] = std::string::npos; 43 } 44 nextPos = next[nextPos]; 45 } 46 } 47 48 for (auto itr = tmpRel.begin(); itr != tmpRel.end(); ++itr) { 49 rel.push_back(*itr); 50 } 51 52 return rel; 53 } 54 55 Solution::~Solution() { 56 delete [] next; 57 }
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];
每次向右移动1位字符，相当于字符串对应的int值左移2位，再将其最低2位置为新的字符的编码值，最后将高2位置0。例如
src CAAAAAAAAAC
subStr CAAAAAAAAA
int 0100000000
subStr AAAAAAAAAC
int 0000000001
字符串遍历O(n)，hash tableO(1)；总时间复杂度O(n)
1 #include <string> 2 #include <vector> 3 #include <unordered_set> 4 #include <cstring> 5 6 bool hashMap[1024*1024]; 7 8 class Solution { 9 public: 10 std::vector<std::string> findRepeatedDnaSequences(std::string s); 11 }; 12 13 std::vector<std::string> Solution::findRepeatedDnaSequences(std::string s) { 14 std::vector<std::string> rel; 15 if (s.length() <= 10) { 16 return rel; 17 } 18 19 // map char to code 20 unsigned char convert[26]; 21 convert[0] = 0; // ‘A‘ - ‘A‘ 00 22 convert[2] = 1; // ‘C‘ - ‘A‘ 01 23 convert[6] = 2; // ‘G‘ - ‘A‘ 10 24 convert[19] = 3; // ‘T‘ - ‘A‘ 11 25 26 // initial process 27 // as ten length string 28 memset(hashMap, false, sizeof(hashMap)); 29 30 int hashValue = 0; 31 32 for (int pos = 0; pos < 10; ++pos) { 33 hashValue <<= 2; 34 hashValue |= convert[s[pos] - ‘A‘]; 35 } 36 37 hashMap[hashValue] = true; 38 39 std::unordered_set<int> strHashValue; 40 41 // 42 for (int pos = 10; pos < s.length(); ++pos) { 43 hashValue <<= 2; 44 hashValue |= convert[s[pos] - ‘A‘]; 45 hashValue &= ~(0x300000); 46 47 if (hashMap[hashValue]) { 48 if (strHashValue.find(hashValue) == strHashValue.end()) { 49 rel.push_back(s.substr(pos - 9, 10)); 50 strHashValue.insert(hashValue); 51 } 52 } else { 53 hashMap[hashValue] = true; 54 } 55 } 56 57 return rel; 58 }
Leetcode：Repeated DNA Sequences详细题解
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原文地址：http://www.cnblogs.com/hzhesi/p/4285793.html
