[Niuke] length of the last word of HJ1 string

Three lines of code to do a problem HJ1 The length of the last word in the string

I mean it doesn't include fixed code ！

🧸 Reading questions

Output several words , Space off , Output the length of the last word .

🧸 Code

Write the final code directly

#include <iostream>
using namespace std;
int main()
{
    string s;
    getline(cin,s);
    cout<<s.size()-s.rfind(' ')-1<<endl;
    return 0;
}

🧸 Decoding code

string s;
establish string
getline(cin,s);
Receive string consecutively
cout<<s.size()-s.rfind(' ')-1<<endl;
use size Subtract the last one ‘ ’( Space ) The location of subtracting 1, because size Is the next position of the last character
So the left open and right closed subtraction is The number of characters .

🧸 Code iteration process

first generation

This is the record from a few months ago

#include <iostream>
using namespace std;
 
int main()
{
    string s;
    //cin>>s;//cin Read space or end of newline  scanf Empathy 
    // Method 1 ： One character, one character 
//     char ch = getchar();
//     //char ch = cin.get();
//     while(ch!='\n')
//     {
//         s+=ch;
//         ch = getchar();
//     }
    // Mode two ：
    getline(cin,s);
    size_t pos = s.rfind(' ');
    if(pos == string::npos)
    {
        cout <<s.size()<<endl;
    }
    else{
        cout << s.size() - pos-1;
    }
    return 0;   
}

Second generation

Today, a few months later, I did it again

#include <iostream>
using namespace std;

int main()
{
    string s;
    getline(cin,s);
    size_t pos = s.rfind(' ');
    int count  = 0;
    while(pos != s.size()-1)
    {
        ++pos;
        ++count;
    }
    cout <<count<<endl;
    return 0;
}

The third generation

When I finish writing , Read the first version of the code , So I think while Circular doing , It's redundant ！ With further transformation

#include <iostream>
using namespace std;

int main()
{
    string s;
    getline(cin,s);
    size_t pos = s.rfind(' ');
    cout<<s.size()-pos-1<<endl;

    return 0;
}

The final version

It turned out pos What you do can also be omitted ！！！

#include <iostream>
using namespace std;
 
int main()
{
    string s;
    getline(cin,s);
    cout<<s.size()-s.rfind(' ')-1<<endl;
    return 0;
}

Other big guys solve it

Count the last word size
That's direct loop reception , Overwrite the previous words , Direct output size, Bullfrog ！

int main() {
    string s;
    while(cin >> s);
    cout << s.size();
    return 0;
}

come on. , I wish you get your favorite offer！