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.
原题链接：https://oj.leetcode.com/problems/repeated-dna-sequences/
straight-forward method（TLE）
算法分析
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
case1:
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
case2：
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]]
case3：
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
时间复杂度
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)
代码实现
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 }
hash table plus bit manipulation method
（view the Show Tags and Runtime 10ms !）
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];
遍历字符串的设计
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
时间复杂度
字符串遍历O(n)，hash tableO(1)；总时间复杂度O(n)
代码实现
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 }