如何推回向量的向量？ [英] how to pushback on a vector of vectors?

问题描述

我正在进行20行输入。我想用一个空格分隔每行的内容，并将其放入向量的向量。如何做一个向量的向量？

 <$> 
 c $ c>玛丽有一只小羊
 lalala山上
太阳升起
  

向量应该看起来像这样。

  ROW 0：{Mary，had a，little，lamb} 
 ROW 1：{lalala，up，the，hill} 
  

这是我的代码....

  
 vector< vector< string> >大; 
 string buf; 
 for（int i = 0; i< 20; i ++）{
 getline（cin，line）; 
 stringstream ss（line）; 
 
 while（ss>> buf）{
（big [i]）。push_back（buf）; 
} 
} 
  

解决方案

代码是正确的，但是你的向量中有零个元素，所以你不能访问 big [i] 。

循环之前的向量大小，在构造函数中，或者像这样：

  big.resize（ruleNum）; 
  

或者，您可以在每个循环步骤中推一个空向量：

  big.push_back（vector< string>（））; 
  

您不需要 big [i] 。

I am taking 20 lines of input. I want to separate the contents of each line by a space and put it into a vector of vectors. How do I make a vector of vectors? I am having have struggles pushing it back...

My input file:

Mary had a little lamb
lalala up the hill
the sun is up

The vector should look like something like this.

ROW 0: {"Mary","had", "a","little","lamb"}
ROW 1: {"lalala","up","the","hill"}

This is my code....

string line; 
vector <vector<string> > big;
string buf;
for (int i = 0; i < 20; i++){
    getline(cin, line);
    stringstream ss(line);

    while (ss >> buf){
        (big[i]).push_back(buf);
    }
}

解决方案

The code is right, but your vector has zero elements in it so you cannot access big[i].

Set the vector size before the loop, either in the constructor or like this:

big.resize(ruleNum);

Alternatively you can push an empty vector in each loop step:

big.push_back( vector<string>() );

You don't need the parentheses around big[i] either.

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