11

我正在使用以下代码来查找类型中的std::vector字符串string。但是如何返回特定元素的位置呢?

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

4

4 回答 4

24

您可以std::distance为此使用:

auto pos = std::distance(vec.begin(), it);

对于 an std::vector::iterator,您还可以使用算术:

auto pos = it - vec.begin();
于 2013-10-18T04:59:19.323 回答
3

使用以下:

if(it != vec.end())
   std::cout<< "Found At :" <<  (it-vec.begin())  ;
于 2013-10-18T04:59:58.573 回答
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;
于 2015-07-17T19:19:21.910 回答
0

使用此语句:

it = find(vec.begin(), vec.end(), "r") - vec.begin();
于 2017-05-08T12:43:58.310 回答