首先按列填充一个锯齿状的二维数组 [英] Filling a jagged 2d array first by columns

问题描述

我想编写一个函数，该函数需要一个2d数组，并用 1 ... n 填充它，但首先要计算列数而不是行数:

I want to write a function that takes an 2d array and fills it with 1...n but counting the columns first instead of the rows:

input = {{0, 0, 0, 0}, {0}, {0}, {0, 0}};

the output should be: {{1, 5, 7, 8}, {2}, {3}, {4, 6}};

如果我要遍历行，然后获得列，我会得到:

if i were to loop through rows and then colums i get:

private static void fill1(int[][] input) {
    int count = 1;
    for (int i = 0; i < input.length; i++) {
        for (int j = 0; j < input[i].length; j++) {
            input[i][j] = count;
            count++;
        }
    }
}

我如何首先遍历各栏?

推荐答案

要首先遍历锯齿状2d数组的列以填充它，您必须事先知道最大列数，但是如果不知道这样，您可以迭代到 Integer.MAX_VALUE ，并在每一步检查列是否仍然存在:

To iterate first over the columns of a jagged 2d array to fill it, you have to know the maximum number of columns beforehand, but if you don't know that, you can iterate to the Integer.MAX_VALUE and check at each step if the columns are still present or not:

int[][] arr = {{0, 0, 0, 0}, {0}, {0}, {0, 0}};

int count = 1;
for (int col = 0; col < Integer.MAX_VALUE; col++) {
    boolean max = true;
    for (int row = 0; row < arr.length; row++) {
        if (col < arr[row].length) {
            arr[row][col] = count;
            count++;
            max = false;
        }
    }
    if (max) break;
}

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

输出:

[1, 5, 7, 8]
[2]
[3]
[4, 6]

---

^{另请参见:如何在不使用存储阵列的情况下将阵列旋转90度?}

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