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原题地址：
https://oj.leetcode.com/problems/repeated-dna-sequences/
题目内容：
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"].
方法：
大概的方向是，遍历所有长度为10的子串，用一个hash表记录每个不同子串的出现次数，最后输出满足条件的子串。
关键问题是：如何更快
我的办法：将A、C、G、T映射到1、2、3、4，然后换算成大整数。为了方便计算，字符串的最左边是最低位。这么说有些语焉不详，举几个例子：
AACG = 4311
CGTA = 1432
然后计算出首个字符串的整数值并加入map，这样，每下一个子串都可以通过 整数值/10 + 下一个字符乘以十亿来得到。
这样，hash值的计算从字符串变成了整数，同时，获得下一个字符串的行为也可以在更快的常数次时间内完成，因为操作字符串的时间开支。
全部代码：
class Solution {
public:
vector<string> findRepeatedDnaSequences(string s) {
unordered_map<long long,int> dict;
unordered_map<long long,int> :: iterator it;
vector<string> res;
long long flag = 1000000000;
if (s.size() <= 10)
return res;
long long num = generateFirstNum(s);
dict[num] = 1;
for (int i = 10; i < s.size(); i ++) {
num /= 10;
long long now = getCharNum(s[i]);
num += now * flag;
it = dict.find(num);
if (it == dict.end()) {
dict[num] = 1;
} else {
dict[num] += 1;
}
}
for (it = dict.begin(); it != dict.end(); it ++) {
if (it->second > 1) {
generateRes(res,it->first);
}
}
return res;
}
long long generateFirstNum(string s) {
long long res = 0;
long long power = 1;
for (int i = 0; i < 10; i ++) {
long long num = getCharNum(s[i]);
res += num * power;
power *= 10;
}
return res;
}
long long getCharNum(char s) {
switch (s) {
case ‘A‘ : return 1;
case ‘C‘ : return 2;
case ‘G‘ : return 3;
case ‘T‘ : return 4;
}
}
char getNumChar(long long s) {
switch (s) {
case 1 : return ‘A‘;
case 2 : return ‘C‘;
case 3 : return ‘G‘;
case 4 : return ‘T‘;
}
}
void generateRes(vector<string> &res,long long target) {
string s;
while (target > 0) {
char now = getNumChar(target % 10);
s = s + now;
target /= 10;
}
res.push_back(s);
}
};
【原创】leetCodeOj --- Fraction to Recurring Decimal 解题报告
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原文地址：http://www.cnblogs.com/shadowmydx/p/4292614.html
