正确实施2D矢量人群
Correct implementation of 2d vector population
本文关键字:2D 更新时间:2023-10-16
我在实现文件中有一个构造函数和方法:
Boggle::Boggle(std::string boardString){
dim = sqrt(boardString.size());
vector<vector<char> > grid(dim, vector<char>(dim));
int co = 0;
for (int i = 0; i < dim; i++)
{
for (int j = 0; j < dim; j++)
{
grid[i][j] = boardString[co];
co++;
}
}
}
void Boggle::printMe() {
for (auto inner : grid)
{
for (auto item : inner)
{
cout << item << " ";
}
cout << endl;
}
}
该程序执行,但无能为力。如您所见,当我声明矢量时,我已经对矢量进行了尺寸。我相信这个问题在于我的逻辑,即从字符串中将角色分配给向量。
在注释中暗示您的vector grid
是您函数的本地。您大多可能想使用类变量,但最终创建了局部变量。您可以使用resize
设置网格的尺寸。同样最好是ceil
sqrt
,以确保我们不会缺少任何字符。
示例:
#include <stdio.h>
#include <vector>
#include <string>
#include <cmath>
#include <iostream>
using namespace std; // Avoid this
class Boggle{
public:
int dim;
vector<vector<char>> grid;
Boggle(string boardString);
void printMe();
};
Boggle::Boggle (std::string boardString)
{
dim = ceil(sqrt(boardString.size ()));
grid.resize(dim, vector <char>(dim));
int co = 0;
for (int i = 0; i < dim; i++) {
for (int j = 0; j < dim; j++)
{
grid[i][j] = boardString[co];
co++;
}
}
}
void Boggle::printMe ()
{
for (auto inner:grid) {
for (auto item:inner)
{
cout << item << " ";
}
cout << endl;
}
}
int main(){
Boggle boggle("hello world");
boggle.printMe();
return 0;
}
结果:
h e l l
o w o
r l d
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