我在这里有一个数组strings
这是我的拼图板视图,
现在一切正常但数组是硬编码的,我需要从0-4
随机生成字符串。
我试图获得random char
并将其设为string
,但这并不起作用。任何提示都会很好。
Random rand = new Random();
char c = (char)(rand.nextInt(5) + '0');
StringBuilder sb = new StringBuilder();
sb.append(c);
String[] debug_board_state = new String[7];
debug_board_state[0] = "0,3,0,0,3,0,2";
debug_board_state[1] = "1,0,2,0,0,1,2";
debug_board_state[2] = "0,2,0,0,0,0,0";
debug_board_state[3] = "0,0,3,0,3,0,4";
debug_board_state[4] = "2,0,0,0,0,1,0";
debug_board_state[5] = "0,1,0,0,1,0,2";
debug_board_state[6] = "2,0,3,0,0,2,0";
更新。
感谢用户回答我能够获得随机矩阵,虽然我遇到了另一个问题,但我需要对矩阵做更多的事情,所以我不想打印出来。这是代码
static private final int WIDTH_EASY = 7;
protected void InitializeEasy() {
Random rand = new Random();
String[][] debug_board_state = new String[7][7];
for (int row = 0; row < debug_board_state.length; row++) {
for (int column = 0; column < debug_board_state[row].length; column++) {
debug_board_state[row][column] = String.valueOf(rand.nextInt(5));
}
}
for (int row = 0; row < debug_board_state.length; row++) {
for (int column = 0; column < debug_board_state[row].length; column++) {
System.out.print(debug_board_state[row][column] + " ");
}
};
for (int i = 0; i < WIDTH_EASY; ++i) {
StringTokenizer tokenizer = new StringTokenizer (debug_board_state[i][i], ",");
int column = 0;
while(tokenizer.hasMoreTokens()) {
String token = tokenizer.nextToken();
getCurrentState().board_elements[i][column] = new BoardElement();
getCurrentState().board_elements[i][column].max_connecting_bridges = Integer.parseInt(token);
getCurrentState().board_elements[i][column].row = i;
getCurrentState().board_elements[i][column].col = column;
if (getCurrentState().board_elements[i][column].max_connecting_bridges > 0) {
getCurrentState().board_elements[i][column].is_island = true;
}
++column;
}
}
}
字符串Tokenizer适用于1d数组但不适用于2d,我需要一些与StringTokenizer相同的东西并将其应用于矩阵。我收到以下错误
java.lang.NullPointerException: Attempt to read from field Island_and_Bridges.Hashi.BoardElement[][] Island_and_Bridges.Hashi.BoardState$State.board_elements on a null object reference
答案 0 :(得分:1)
虽然我认为int[][]
是一个更好的主意,但这里是String[][]
解决方案。您可以使用String.valueOf(rand.nextInt(5))
在矩阵中生成元素:
import java.util.Random;
public class Main {
public static void main(String[] args) {
Random rand = new Random();
String[][] matrix = new String[7][7];
for (int row = 0; row < matrix.length; row++) {
for (int column = 0; column < matrix[row].length; column++) {
matrix[row][column] = String.valueOf(rand.nextInt(5));
}
}
for (int row = 0; row < matrix.length; row++) {
for (int column = 0; column < matrix[row].length; column++) {
System.out.print(matrix[row][column] + " ");
}
System.out.println();
}
}
}
更新
for (int row = 0; row < WIDTH_EASY; ++row) {
for (int column = 0; column < WIDTH_EASY; ++column) {
getCurrentState().board_elements[row][column] = new BoardElement();
getCurrentState().board_elements[row][column].max_connecting_bridges = Integer.parseInt(debug_board_state[row][column]);
getCurrentState().board_elements[row][column].row = row;
getCurrentState().board_elements[row][column].col = column;
if (getCurrentState().board_elements[row][column].max_connecting_bridges > 0) {
getCurrentState().board_elements[row][column].is_island = true;
}
}
}
答案 1 :(得分:0)
这样的东西?
伪代码:
String[][] debug_board_state = new String[7][7];
for (int x = 0; x < debug_board_state.size(); x++) {
for (int y = 0; y < debug_board_state[x].size(); y++) {
debug_board_state[x][y] = new_random_character();
}
}
答案 2 :(得分:0)
0-4位于ASCII范围内的49到52之间:
Random rand = new Random();
char c = (char)(rand.nextInt(4)+49);
StringBuilder sb = new StringBuilder();
sb.append(c+'0');
答案 3 :(得分:0)
也许,你想要这样的东西:
public void initBoard() {
Random random = new Random();
String[][] board = new String[7][7];
for (int i=0; i < board.size(); i++) {
for (int j=0; j < board[].size(); j++) {
board[i][j] = String.valueOf(random.nextInt() % 5);
}
}
}
它将使用随机数String
初始化您的电路板。