创建初始重复数据的二维字符串数组的最有效方法是什么？

Question

在 Python 中，通常要定义所述游戏板，我会使用以下代码：

board = [['0','0','0','0','0','0','0','0','0','0'] for i in range(10)]

但是，到目前为止，我能够找到产生相同结果的唯一方法是使用 Java：

public static String[][] board = {
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"},
        {"0","0","0","0","0","0","0","0","0","0"}};

当然必须有更有效的方法来做到这一点。我知道如果它是整数，我可以有效地做到这一点，但考虑到这种情况，它必须是字符串。我原以为 for 循环也可以应用，但我似乎找不到合适的语法；也许它不适用于Java。有什么建议吗？

Answer 1

您可以遍历每一行并使用Arrays.fill：

public static String[][] board = new String[rows][cols];
static {
    for (String[] row : board) Arrays.fill(row, "0");
}

Demo

Answer 2

您可以使用嵌套的for 循环来创建 10 行和 10 列。因此，在第一个循环（创建一个“行”）中，您需要添加另一个循环来填充“列”。

String[][] board = new String[10][10];

// 10 rows
for (int i = 0; i < 10; i++) {
    // 10 columns
    for (int j = 0; j < 10; j++) {
        // Set the value of this row/column
        board[i][j] = "0";
    }
}

这显然不如 Python 的单行解决方案优雅，但它可以满足您的需求。

还有其他方法可以做到这一点（例如在每个循环中将一个完整的 10 位数组添加到 board 数组），但我发现这对于可扩展性和可读性来说是最好的。

---

为了更清楚起见，如果您将来可能需要更改行数或列数，最好将这些值移动到变量中并改用它们：

int rows = 10;
int cols = 10;

String[][] board = new String[rows][cols];

for (int i = 0; i < rows; i++) {
    for (int j = 0; j < cols; i++) {
        board[i][j] = "0";
    }
}

这样你只需要更新变量而不是修改for循环和数组声明。

Answer 3

你可以这样做。因为Strings 是不可变的，所以在这里使用clone() 是安全的。感谢@iota 推荐setAll

int n = 10;
String[][] arr = new String[n][n];
Arrays.fill(arr[0], "0");
Arrays.setAll(arr, i->arr[0].clone());


for(String[] a : arr) {
    System.out.println(Arrays.toString(a));
}

打印

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Answer 4

以下是基于stream 的解决方案，该解决方案使用函数式样式而非迭代式样式：

return Stream.generate(() -> 
                Collections.nCopies(10, "0").toArray(String[]::new))
        .limit(10)
        .toArray(String[][]::new);

请注意，根据可读性等要求，可能会认为经典的迭代样式更可取。

说明

Collections.nCopies(10, "0") 创建一个包含 10 个 "0" 值的 List<String>。使用Collection.toArray(IntFunction) 将其转换为String 数组。这需要一个IntFunction，它会生成一个给定长度的数组。 String[]::new 是一个 constructor lambda，它实现了 IntFunction<String[]>，创建了一个给定长度的数组。

Stream.generate 通过调用给定的Supplier 创建每个元素的无限流。在这种情况下，供应商创建了前面提到的 10 元素长列表，结果是 Stream<String[]>。使用了生成器，而不是 Collections.nCopies，因为 String[][] 中的每个 String[] 都需要是不同的对象。

无限流仅限于使用Stream.limit(10) 的十个元素。

最后，使用Stream.toArray(IntFunction) 将Stream<String[]> 收集到String[][] 中，其工作方式与已经描述的Collection.toArray 方法相同。

Answer 5

这是 for 循环和手动创建一行单元格的混合体。

String[][] board = new String[10][];
for (int i = 0; i < 10; i++) {
    board[i] = new String[]{"0","0","0","0","0","0","0","0","0","0"};
}

Answer 6

String#repeat 从 11 开始

String[][] board = new String[10][];

1. Arrays.setAll - 每行重复 0 十次并拆分为一个数组：

   Arrays.setAll(board, i -> "0".repeat(10).split(""));
   

2. Arrays.fill - 在这种情况下，结果是一样的：

   Arrays.fill(board, "0".repeat(10).split(""));
   

输出：

Arrays.stream(board).map(Arrays::toString).forEach(System.out::println);

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

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^{另见：How can I fill this 2D array with X characters?}