java中的2D数组类,带有环绕边

时间:2012-01-30 00:49:25

标签: java arrays arraylist

我在Java中创建一个带有环绕边缘的2D数组类。因此,在10x20数组中,条目(2,-1)与(2,19)相同,(4,22)与(4,2)相同。我还希望能够将此结构作为“扁平化阵列”访问,因此我将使用类似一维阵列的结构来存储对象。我会使用一维数组,但后来我不能使用泛型类型,所以我使用的是ArrayList。该课程尚未完成,但我发布了以下代码。

我的第一个问题是算法:在下面的代码中,我使用modRow(i)和modCol(j)函数访问元素(i,j)。它工作正常,但我想知道是否有人能想到更好的方法来做到这一点。然后get(i,j)函数调用这些函数。我希望能够尽快访问元素。

我的第二个问题是针对Java的:我对Java很新,并且正在从C ++项目中提取这些代码。我有一个set()函数,它允许我在数组中的位置i处设置一个元素(存储在ArrayList中)。但是,set函数仅在已调用add()函数并将该元素设置在flattened out数组中的位置i时才有效。如果有人对如何让这个班级更好地工作有任何建议,我会很高兴见到他们。

import java.util.*;

public class Grid<T> {

    ArrayList<T> array; // holds objects in grid - would rather use an array
                        // but that won't support generics
    int rows;  // number of rows in grid
    int cols;  // number of cols in grid
    int length; // total number of objects in grid - maybe just get this from ArrayList
    int conNum; // number of connections for each point on grid (either 4 or 8) 
                // used for determining distances between points on the grid

    // constructor
    Grid(int row, int col, int numCon) {
        rows = row;
        cols = col;
        length = rows * cols;
        conNum = numCon;
        array = new ArrayList<T>(length);
    }

    // returns total size of grid
    public int len() {
        return length;
    }

    // returns number of rows
    public int row() {
        return rows;
    }

    // returns number of columns
    public int col() {
        return cols;
    }

    // wish I didn't have to use this 
    // would be nice to just be able to use set(int i, T t)
    public void add(T t) {
        array.add(t);
    }

    // sets object i in flattened out array
    public void set(int i, T t) {
        array.set(i, t);
    }

    // returns object i in flattened out array
    // for faster access when user just needs to iterate through all objects 
    // in grid without respect to position in 2D grid
    public T get(int i) {
        return array.get(i);
    }

    // returns the row position of i in grid - adjusted for warp around edges 
    // is there a better way to do this?  
    private int modRow(int i) {
        if(i >=0) {
            if(i < rows) {
                return i;
            } else { // i >= rows
                return i % rows;
            }
        } else { // i < 0
            return (i % rows + rows) % rows;
        }
    }

    // returns the column position of j in grid - adjusted for warp around edges 
    // is there a better way to do this? 
    private int modCol(int j) {
        if(j >=0) {
            if(j < cols) {
                return j;
            } else { // j >= cols
                return j % cols;
            }
        } else { // j < 0
            return (j % cols + cols) % cols;
        }
    }

    // sets object at (i,j) value from store adjusted for wrap around edges 
    public void set(int i, int j, T t) {
        array.set(modRow(i) * cols + modCol(j), t);
    }

    // gets object at (i,j) value from store adjusted for wrap around edges 
    public T get(int i, int j) {
        return array.get(modRow(i) * cols + modCol(j));
    }

    // returns distance on the grid between two objects at (y1,x1) and (y2,x2)
    public int dist(int y1, int x1, int y2, int x2) {
        int y = distFirst(y1, y2);
        int x = distSecond(x1, x2);
        if (conNum == 4) // taxicab distance
        {
            return y + x;
        } else { //if(conNum == 8) supremum distance
            return Math.max(y, x);
        }
    }

    // returns distance on the grid between the first coordinates y1 & y2 of two objects
    public int distFirst(int y1, int y2) {
        int dist = Math.abs(modRow(y2) - modRow(y1));
        return Math.min(dist, rows - dist);
    }

    // returns distance on the grid between the second coordinates x1 & x2 of two objects
    public int distSecond(int x1, int x2) {
        int dist = Math.abs(modCol(x2) - modCol(x1));
        return Math.min(dist, cols - dist);
    }
}

2 个答案:

答案 0 :(得分:1)

不确定是什么导致你也认为你不能将泛型用于数组,因为你可以。因此,请将内部存储定义和调用集更改为您的内容。

您的modRows / modCols函数可以(稍微)更高效。索引只能在一个方向上超出范围,因此您可以摆脱嵌套的if检查。

if(index < 0) return index + max_size;
else if(index >= max_size) return index % max_size;
else return index;

答案 1 :(得分:0)

@Perception:原始index < 0支票有什么问题?

if(index < 0) return (index % max_size + max_size) % max_size;    // <-- like that
else if(index >= max_size) return index % max_size;
else return index;