我使用以下代码查找std::vector
string
类型的字符串。但是如何返回特定元素的位置?
代码:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
vector<string> vec;
vector<string>::iterator it;
vec.push_back("H");
vec.push_back("i");
vec.push_back("g");
vec.push_back("h");
vec.push_back("l");
vec.push_back("a");
vec.push_back("n");
vec.push_back("d");
vec.push_back("e");
vec.push_back("r");
it=find(vec.begin(),vec.end(),"r");
//it++;
if(it!=vec.end()){
cout<<"FOUND AT : "<<*it<<endl;
}
else{
cout<<"NOT FOUND"<<endl;
}
return 0;
}
输出:
FOUND AT : r
预期输出:
FOUND AT : 9
答案 0 :(得分:23)
您可以使用std::distance
:
auto pos = std::distance(vec.begin(), it);
对于std::vector::iterator
,您还可以使用算术:
auto pos = it - vec.begin();
答案 1 :(得分:3)
使用以下内容:
if(it != vec.end())
std::cout<< "Found At :" << (it-vec.begin()) ;
答案 2 :(得分:3)
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
vector<string> vec;
vector<string>::iterator it;
vec.push_back("H");
vec.push_back("i");
vec.push_back("g");
vec.push_back("h");
vec.push_back("l");
vec.push_back("a");
vec.push_back("n");
vec.push_back("d");
vec.push_back("e");
vec.push_back("r");
it=find(vec.begin(),vec.end(),"a");
//it++;
int pos = distance(vec.begin(), it);
if(it!=vec.end()){
cout<<"FOUND "<< *it<<" at position: "<<pos<<endl;
}
else{
cout<<"NOT FOUND"<<endl;
}
return 0;