从数字和字母表中快速生成大量随机字符串

Question

我必须从"az","AZ"和"0-9"随机生成一个大的字符串(10k甚至更多),其大小为32个字符.

到目前为止,我的脑海中有以下代码(O(N*32)),但我想知道是否有更好的方法可以做到这一点.

int N = 10000;           
vector<string> vecStr;

for (int index=0; index<N; index++)
{
  string str;
  for (int i = 0; i < 32; ++i)
  {
    int randomChar = rand()%(26+26+10);        
    if (randomChar < 26)
      str += 'a' + randomChar;
    else if (randomChar < 26+26)
      str += 'A' + randomChar - 26;
    else
      str += '0' + randomChar - 26 - 26;
  }
  vecStr.push_back(str);
} 

Answer 1

你不会找到比O(N*len)更好的解决方案,其中N是字符串的数量,len是其中每个的长度.也就是说,在某个地方,我确信有一个玷污的贴纸,我可以通过编写最密集的代码来获得这样做:

#include <iostream>
#include <iterator>
#include <vector>
#include <random>
#include <algorithm>

int main()
{
    static const char alphabet[] =
        "abcdefghijklmnopqrstuvwxyz"
        "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        "0123456789";

    static const size_t N_STRS = 10000;
    static const size_t S_LEN = 32;

    std::random_device rd;
    std::default_random_engine rng(rd());
    std::uniform_int_distribution<> dist(0,sizeof(alphabet)/sizeof(*alphabet)-2);

    std::vector<std::string> strs;
    strs.reserve(N_STRS);
    std::generate_n(std::back_inserter(strs), strs.capacity(),
        [&] { std::string str; 
              str.reserve(S_LEN); 
              std::generate_n(std::back_inserter(str), S_LEN,
                   [&]() { return alphabet[dist(rng)];}); 
              return str; });
    std::copy(strs.begin(), strs.end(), std::ostream_iterator<std::string>(std::cout, "\n"));
    return 0;
}

输出(为简洁省略了9990行= P)

MRdeOWckfKy8GTFt0YmQMcM6SABJc934
XvdcatVsv6N9c1PzQGFFY6ZP943yIrUY
xpHzxUUyAizB6BfKldQzoePrm82PF1bn
kMUyPbflxk3yj3IToTFqYWnDq6aznKas
Ey0W5SF37VaeEY6PxWsBoxlNZTv9lOUn
iTx7jFRTHHW6TfYl7N3Hne4yu7kgAzp5
0ZamlaopjLyEvJbr6fzJPdXmjLOohtKh
6ZYeqj47nCMYKj0sCGl2IHm28FmvuH8h
oTDYRIA1trN1A2pQjsBwG3j9llzKIMhw
5zlpvSgTeLQ38eFWeSDoSY9IHEMHyzix

请注意,您可能会惊讶地发现它的速度有多快.引擎盖下发生了很多事情.最后,这使用了C++ 11随机库,特别是均匀分布,这消除了传统rand() % n解决方案通常遇到的模偏差n.