[英]Placing numbers in a 2D array by a chess knight move pattern in Java using recursion
public class ChessComplete
{
private int size;
private int[][] board;
private long callNum;
public ChessComplete(int size)//constructor with 2D array size as a parameter
{
this.size = size;
board = new int [size][size];
board[0][0] = 1;
}
public boolean isValid(int r, int c)//To check the if the number that is added is in the 2D array is in bound
{
if(r < 0 || c < 0 || r >= size || c >= size)
{
return false;
}
return true;
}
/*
* Move through the 2D array by placing the consecutive number for each (row,col) until its full
* Moves Are only allowed in a chess knight pattern
*/
public boolean move(int r, int c, int count) {
callNum++;
if (!isValid(r, c)) {
return false;
}
if (count == (size * size))// Base case if count reaches the end of 2D
// array
{
return true;
}
board[r][c] = count;// fills the positon with the current count
if (board[r][c] == 0) {
return false;
}
// Check if each possible move is valid or not
if (board[r][c] != 0) {
for (int i = 0; i < 8; i++) {
int X[] = { 2, 1, -1, -2, -2, -1, 1, 2 };
int Y[] = { 1, 2, 2, 1, -1, -2, -2, -1 };
// Position of knight after move
int x = r + X[i];
int y = c + Y[i];
if (move(x, y, count + 1)) {
return move(x, y, count + 1);
}
}
}
board[r][c] = 0;
return false;
}
public long getSteps()//Number of reccursive trials
{
return callNum;
}
public void displayBoard()
{
String s = " ";
for(int r = 0; r < size; r++)
{
for(int c = 0; c < size; c++)
{
System.out.print(board[r][c] + " ");
}
System.out.println();
}
}
}
輸出為:
1, 0, 0, 0, 0,
0, 0, 0, 23, 0,
0, 2, 0, 0, 0,
0, 0, 0, 0, 24,
0, 0, 3, 0, 0
Recursive method call count: 78,293,671
注意,在位置(行,列) (0, 0)
有一個1
,在位置(2, 1)
有2
。 如您所見,棋盤中的騎士以類似的方式移動。 這樣,我們需要用連續的數字填充整個2D數組,以使其嚴格遵循騎士模式。
我不明白為什么整個2D數組沒有被所有其他連續數字填充。 例如,在2D數組中用3填充位置后,數字會一直跳到23。
在執行移動之前,您無需檢查正方形是否已被占用,因此您的解決方案包括重復寫入的重復移動。
更改isValid
以檢查目標正方形是否為空:
public boolean isValid(int r, int c) {
if (r < 0 || c < 0 || r >= size || c >= size || board[r][c] != 0) {
return false;
}
return true;
}
...並刪除初始化步驟board[0][0] = 1;
在構造函數中(應該通過對move
的第一個調用來設置)。
另外(但不是致命的),
if (move(x, y, count + 1)) {
return move(x, y, count + 1);
}
應該
if (move(x, y, count + 1)) {
return true;
}
並且檢查if (board[r][c] == 0)
和if (board[r][c] != 0)
是否不執行任何操作,因為它們是在設置board[r][c] = count;
。
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