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我有一个 2D 环境，我想在其中获取摩尔附近的一些单元格（东、西、北和东单元格）。有时一个单元格只能有 3 或 2 个邻居（如果我们从边界或网格的一角检查）。

在此之上，我想区分没有边界的情况（环面世界，上升的东西再次出现......）。在这种情况下，每个单元格恰好有 4 个邻居。

因此，我制作了该算法的 2 个版本，一个使用 std::vector 坐标，另一个使用 std::array 可选坐标（我将坐标称为一对短整数）。见下文 ：

带有成对向量的版本

std::vector<std::pair<uint16_t, uint16_t>>* Environment::getMooreNeighborhood(uint16_t x, uint16_t y) const{
    std::vector<std::pair<uint16_t, uint16_t>>* neighborhood = new std::vector<std::pair<uint16_t, uint16_t>>();
    std::pair<uint16_t, uint16_t> west_cell(x - 1, y);
    std::pair<uint16_t, uint16_t> east_cell(x + 1, y);
    std::pair<uint16_t, uint16_t> north_cell(x, y - 1);
    std::pair<uint16_t, uint16_t> south_cell(x, y + 1);

    if(this->torus){
        if(x == this->width - 1){
            east_cell.first = 0;
        }
        else if(x == 0){
            west_cell.first = this->width - 1;
        }
        if(y == 0){
            north_cell.second = this->height - 1;
        }
        else if(y == this->height - 1){
            south_cell.second = 0;
        }
        neighborhood->push_back(east_cell);
        neighborhood->push_back(west_cell);
        neighborhood->push_back(south_cell);
        neighborhood->push_back(north_cell);
    }else{
        if(x > 0){
            neighborhood->push_back(east_cell);
        }
        if(x < this->width - 1){
            neighborhood->push_back(west_cell);
        }
        if(y > 0){
            neighborhood->push_back(north_cell);
        }
        if(y < this->height - 1){
            neighborhood->push_back(south_cell);
        }
    }
    return neighborhood;
}

带有可选对数组的版本

std::array<std::optional<std::pair<uint16_t,uint16_t>>, 4> Environment::getMooreNeighborhood(uint16_t x, uint16_t y) const {
    std::array<std::optional<std::pair<uint16_t,uint16_t>>, 4> neighborhood;
    std::optional<std::pair<uint16_t, uint16_t>> west_cell;
    std::optional<std::pair<uint16_t, uint16_t>> east_cell;
    std::optional<std::pair<uint16_t, uint16_t>> north_cell;
    std::optional<std::pair<uint16_t, uint16_t>> south_cell;

    if(this->torus){
        if(x == this->width - 1){
            east_cell = std::pair<uint16_t, uint16_t>(0, y);
        }
        else if(x == 0){
            west_cell = std::pair<uint16_t, uint16_t>(this->width - 1, y);
        }
        if(y == 0){
            north_cell = std::pair<uint16_t, uint16_t>(x, this->height - 1);
        }
        else if(y == this->height - 1){
            south_cell = std::pair<uint16_t, uint16_t>(x, 0);
        }
    }else{
        if(x > 0){
            east_cell = std::pair<uint16_t, uint16_t>(x - 1, y);
        }
        if(x < this->width - 1){
            west_cell = std::pair<uint16_t, uint16_t>(x + 1, y);
        }
        if(y > 0){
            north_cell = std::pair<uint16_t, uint16_t>(x, y - 1);
        }
        if(y < this->height - 1){
            north_cell = std::pair<uint16_t, uint16_t>(x, y + 1);
        }
    }
    neighborhood[0] = west_cell;
    neighborhood[1] = east_cell;
    neighborhood[2] = north_cell;
    neighborhood[3] = south_cell;
    return neighborhood;
}

我必须多次执行此操作，而且我不想因为执行不当而自取其辱。一个选择比另一个更好还是我绝对错了，还有另一种有效的方法？

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