I want to replace all occurrence of 'a' with 'b', and 'c' with 'd'.
我想用“b”替换所有出现的'a',用'd'替换'c'。
My current solution is:
我目前的解决方案是:
std::replace(str.begin(), str.end(), 'a', 'b');std::replace(str.begin(), str.end(), 'c', 'd');Is it possible do it in single function using the std?
是否可以使用std在单个函数中执行此操作?
2 个解决方案
#1
1
Tricky solution:
#include <algorithm>#include <string>#include <iostream>#include <map>int main() { char r; //replacement std::map<char, char> rs = { {'a', 'b'}, {'c', 'd'} }; std::string s = "abracadabra"; std::replace_if(s.begin(), s.end(), [&](char c){ return r = rs[c]; }, r); std::cout << s << std::endl;}#2
4
If you do not like two passes, you can do it once:
如果你不喜欢两次通过,你可以做一次:
std::transform(std::begin(s), std::end(s), std::begin(s), [](auto ch) { switch (ch) { case 'a': return 'b'; case 'c': return 'd'; } return ch; });#1
1
Tricky solution:
#include <algorithm>#include <string>#include <iostream>#include <map>int main() { char r; //replacement std::map<char, char> rs = { {'a', 'b'}, {'c', 'd'} }; std::string s = "abracadabra"; std::replace_if(s.begin(), s.end(), [&](char c){ return r = rs[c]; }, r); std::cout << s << std::endl;}#2
4
If you do not like two passes, you can do it once:
如果你不喜欢两次通过,你可以做一次:
std::transform(std::begin(s), std::end(s), std::begin(s), [](auto ch) { switch (ch) { case 'a': return 'b'; case 'c': return 'd'; } return ch; });