英文:
Fill an integer array with a series of number that is a string (java)
问题
# 在下面写你的代码前面的英文说明部分
number = "110034043312132121220023020423340432" # 但可能是任何转换为字符串的大数
print("以字符串形式输入长度: ")
input_str = input() # abcdefg
length = len(input_str) # 这意味着我会得到 6。
# 创建一个二维字符数组
rows = length
columns = len(number)
char_array = [[" " for _ in range(columns)] for _ in range(rows)]
# 填充字符数组
for i in range(rows):
for j in range(columns):
char_array[i][j] = number[j]
# 打印字符数组
for i in range(rows):
for j in range(columns):
print(char_array[i][j], end=" ")
print()
请注意，这只是一个伪代码示例，因为您的原始代码是Java，而您要求使用Python翻译。实际上，数组的索引可能会略有不同。另外，您还需要导入Python的输入函数（input）和适当的打印语句。这只是一个简单示例，您可能需要根据实际情况进行适当的调整。
英文:
Im trying to write the following program, where i have a string that a number is generated possibly like 37 digits.
Is it possible to fill an array that i know the length but not the rows because the string can be anything.
String number; //110034043312132121220023020423340432 but can be any big number that is converted to a string
System.out.println("Give length in a form of a string: ");
String input= scan.nextLine(); //abcdefg
int length = input.length(); //this means i get 6 back.
And save this in a character array where i know the length from input.lenght() and i dont know the rows
A B C D E F G
1 1 0 0 3 4 0
4 3 3 1 2 1 3
2 1 2 1 2 2 0
0 2 3 0 2 0 4
2 3 3 4 0 4 3
2
I know the answer might be something stupid simple but please be coder noob friendly. Thanks
答案1
得分: 0
为了做到这一点，你可以考虑使用一个列表。
ArrayList<int[]> list = new ArrayList<int[]>();
在每一行中，你可以向列表中添加一个新的数组。
如果你希望最终的结果是一个二维数组，你可以将列表很容易地转换为一个数组。
英文:
To do this, you might be able to use a list instead.
ArrayList<int[]> list = new ArrayList<int[]>();
and in each line, you can add a new array into the list.
If you want the final result to be a 2d array, you can easily convert a list to an array.
答案2
得分: 0
public static char[][] createMatrix(String num, String title) {
int totalRows = (int)Math.ceil((double)num.length() / title.length());
char[][] matrix = new char[totalRows + 1][title.length()];
for (int i = 0; i < matrix[0].length; i++)
matrix[0][i] = title.charAt(i);
for (int i = 0, row = 1, col = 0; i < num.length(); i++) {
matrix[row][col++] = num.charAt(i);
if (col == matrix[row].length) {
row++;
col = 0;
}
}
return matrix;
}
英文:
public static char[][] createMatrix(String num, String title) {
int totalRows = (int)Math.ceil((double)num.length() / title.length());
char[][] matrix = new char[totalRows + 1][title.length()];
for (int i = 0; i < matrix[0].length; i++)
matrix[0][i] = title.charAt(i);
for (int i = 0, row = 1, col = 0; i < num.length(); i++) {
matrix[row][col++] = num.charAt(i);
if (col == matrix[row].length) {
row++;
col = 0;
}
}
return matrix;
}
答案3
得分: 0
解决方案可能如下：
1. 使用 `input.length` 作为结果二维数组中列的最大数量。
2. 计算结果二维数组中的行数 `nums / cols`，如果有余数则添加一行额外的行。
3. 结果中的最后一行可能是不规则的：包含少于 `cols` 个元素。
String number = "110034043312132121220023020423340432"; // 但可以是任何转换为字符串的大数字
System.out.println("以字符串形式给出长度：");
String input = "ABCDEFG"; //scan.nextLine(); //abcdefg
int cols = input.length(); // 结果数组中的列数
int nums = number.length(); // 数字的总数
int rows = nums / cols + (nums % cols == 0 ? 0 : 1);
int[][] result = new int[rows][]; // 结果可能是不规则的
for (int i = 0; i < rows - 1; i++) {
result[i] = new int[cols];
}
result[rows - 1] = new int[nums % cols == 0 ? cols : nums % cols];
for (int i = 0; i < nums; i++) {
int r = i / cols;
int c = i % cols;
result[r][c] = number.charAt(i) - '0';
}
// 打印结果
System.out.println(input.replaceAll(".", "$0 ")); // 在列之间打印带有空格的标题
for (int[] row : result) {
System.out.println(Arrays.toString(row)
.replaceAll("[^\\s\\d]", "") // 从输出中删除括号和逗号
);
}
输出：
A B C D E F G
1 1 0 0 3 4 0
4 3 3 1 2 1 3
2 1 2 1 2 2 0
0 2 3 0 2 0 4
2 3 3 4 0 4 3
2
英文:
The solution may be as follows:
- Use
input.lengthas the max number of columns in the resulting 2D array - Calculate the number of rows in the resulting 2D array
nums / cols, adding one extra row if there's a remainder. - The last row in the result may be jagged: contain less than
colselements.
String number = "110034043312132121220023020423340432"; // but can be any big number that is converted to a string
System.out.println("Give length in a form of a string: ");
String input= "ABCDEFG" ; //scan.nextLine(); //abcdefg
int cols = input.length(); // columns in the resulting array
int nums = number.length(); // total count of numbers
int rows = nums / cols + (nums % cols == 0 ? 0 : 1);
int[][] result = new int[rows][]; // the result may be jagged
for (int i = 0; i < rows - 1; i++) {
result[i] = new int[cols];
}
result[rows - 1] = new int[nums % cols == 0 ? cols : nums % cols];
for (int i = 0; i < nums; i++) {
int r = i / cols;
int c = i % cols;
result[r][c] = number.charAt(i) - '0';
}
// print the result
System.out.println(input.replaceAll(".", "$0 ")); // print header with spaces between columns
for (int[] row : result) {
System.out.println(Arrays.toString(row)
.replaceAll("[^\\s\\d]", "") // remove brackets and comma from output
);
}
Output:
A B C D E F G
1 1 0 0 3 4 0
4 3 3 1 2 1 3
2 1 2 1 2 2 0
0 2 3 0 2 0 4
2 3 3 4 0 4 3
2
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