我有关于如何生成范围为1-8的随机数的问题,我只需要2对这些数字而不是1,1,1或3,3,3。这是我已经导入java.util.random的代码:
int[][] displayArray = new int[4][4];
int[][] fieldArray = new int[4][4];
//generates random Values where "i" represents rows and "j" represents columns
for (int i = 0; i < fieldArray.length; i++) {
for (int j = 0; j < fieldArray.length; j++) {
fieldArray[i][j] = (int) (Math.random() * 8); //generates table with randomized numbers.
}
}
//Print the values
for (int i = 0; i < fieldArray.length; i++) {
for (int j = 0; j < fieldArray.length; j++) {
System.out.print("[" + fieldArray[i][j] + "]" + "\t");
}
// Every time we finish printing a row we jump to the next line.
System.out.print("\n");
}
}
答案 0 :(得分:3)
此
for (int j = 0; j < fieldArray.length; j++) {
应该是
for (int j = 0; j < fieldArray[i].length; j++) {
在这两个地方。另外,我更喜欢
Random rand = new Random();
int val = rand.nextInt(8) + 1; // <-- a random number between 1 and 8.
但使用Math.random()
您还需要添加+ 1
,因为您将包含0(并排除8),
fieldArray[i][j] = 1 + (int) (Math.random() * 8);
答案 1 :(得分:2)
您不希望拥有随机数字,而是您的数字的随机顺序。 Collections.shuffle()
提供了一种方法(在List
中随机化顺序)。
此代码适用于任何四边形尺寸(> 0),创建一个填充数字的四边形数组,从1开始,每个数字包含两次,如果可能的话(对于奇数四边形尺寸,您会获得奇数个数字):< / p>
int quadSize = 4; //4x4 = 16 tiles, containing the numbers 1-8 twice
List<Integer> values = new ArrayList<>(quadSize * quadSize);
// add number twice (1,1,2,2, ....)
int num = 0;
boolean first = false;
while (values.size() < quadSize * quadSize) {
if (!first) {
num++;
}
values.add(num);
first = !first;
}
// Shuffle the values (random order)
Collections.shuffle(values);
// fill into two-dimensional array
int[][] fieldArray = new int[quadSize][quadSize];
int index = 0;
for (int value : values) {
fieldArray[index / quadSize][index % quadSize] = value;
index++;
}
使用您的打印方法输出:
[5] [7] [1] [6]
[2] [3] [8] [3]
[6] [8] [5] [1]
[4] [2] [7] [4]