英文:
Printing a snake pattern using an array
问题
以下是翻译好的部分：
//蛇形移动带有数字
import java.util.Scanner;
public class SnakeMove {
public static void main(String[] args) {
//创建 Scanner 对象
Scanner inScan = new Scanner(System.in);
//提示用户从 0 到 16 选择行数
System.out.println("选择行数，范围从 0 到 16。");
//使用 nextInt() 方法从用户获取输入
//使用变量 int row
int row = inScan.nextInt();
//提示用户从 0 到 16 选择列数
System.out.println("选择列数，范围从 0 到 16。");
//使用 nextInt() 方法从用户获取输入
//使用变量 int col
int col = inScan.nextInt();
if (row != col) {
System.out.println("再次运行程序，选择相同的行数和列数。");
System.exit(0);
}
int[][] arr = move(row, col);
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr.length; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}//main 方法
static int[][] move(int row, int col) {
boolean flag = true;
int count = 1;
int[][] array = new int[row][col];
for (int j = 0; j < array[0].length; j++) {
if (flag) {
for (int i = 0; i < array.length; i++) {
//将 count 的增量值赋值给特定的数组单元
array[i][j] = count;
count++;
}
flag = false;
} else {
//逐行递减向上移动
for (int i = array.length - 1; i > 0; i--) {
//将 count 的增量值赋值给特定的数组单元
array[i][j] = count;
count++;
}
flag = true;
}
}//逐列增加
return array;
}//move 方法
}//结束 SnakeMove 类
英文:
I'm having trouble with an assignment where we are required to print out this array:
1 10 11 20 21
2 9 12 19 22
3 8 13 18 23
4 7 14 17 24
5 6 15 16 25
My code is somewhat correct but it is not printing 10 and 19 where it should be.
My output:
Choose a number for the rows from 0 to 16.
5
Choose a number for the columns from 0 to 16
5
1 0 10 0 19
2 9 11 18 20
3 8 12 17 21
4 7 13 16 22
5 6 14 15 23
My code:
//snake move with the number
import java.util.Scanner;
public class SnakeMove {
public static void main(String[] args) {
//create Scanner object
Scanner inScan = new Scanner(System.in);
//prompt the user to choose number for the Row from 0 to 16
System.out.println("Choose a number for the rows from 0 to 16.");
//take the input from user with nextInt() method
//use the variable int row
int row = inScan.nextInt();
//prompt the user to choose number for the Col from 0 to 16
System.out.println("Choose a number for the columns from 0 to 16");
//take the input from user with nextInt()
//use the variable int col
int col = inScan.nextInt();
if (row != col) {
System.out.println("Run the program again and choose the same number for Row and Col");
System.exit(0);
}
int[][] arr = move(row, col);
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr.length; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}//main method
static int[][] move(int row, int col) {
boolean flag = true;
int count = 1;
int[][] array = new int[row][col];
for (int j = 0; j < array[0].length; j++) {
if (flag) {
for (int i = 0; i < array.length; i++) {
//assign the increment value of count
// to specific array cells
array[i][j] = count;
count++;
}
flag = false;
} else {
//row decrement going up
for (int i = array.length - 1; i > 0; i--) {
//assign the increment value of count
// to specific array cells
array[i][j] = count;
count++;
}
flag = true;
}
}//column increment
return array;
}//move method
}//end SnakeMove class
Can anyone detect what is causing the error? Any help would be appreciated.
答案1
得分: 1
我认为这是一个不错的替代方案，容易理解。如果你有同样问题的简化版本，请评论分享。
## 代码：##
import java.util.Arrays;
import java.util.Scanner;
public class SnakePatternProblem {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in); // *输入*
int row = scanner.nextInt();
int col = scanner.nextInt();
int[][] array = new int[row][col];
// 用通用表达式初始化数组的第一行
for (int i = 0; i < col; i++) {
if (i % 2 != 0) array[0][i] = col * (i+1);
else array[0][i] = (col * i) + 1;
}
array[0][0] = 1; // 此元素不在上述循环中涵盖。
// 嵌套循环用于根据第一行元素递增和递减。
for (int i = 1; i < row; i++) {
for (int j = 0; j < col; j++) {
if (j % 2 == 0) array[i][j] = array[i-1][j] + 1;
else array[i][j] = array[i-1][j] - 1;
}
}
System.out.println(Arrays.deepToString(array)); // *输出*
}
}
----------
## 解释：##
考虑一个名为 "arr" 的 *5 x 5* 矩阵的第一行，呈蛇形排列：
1 10 11 20 21
奇数列中的元素等于 ((当前列 + 1) * 总列数)；
例如：arr[0][1] = (1 + 1) * 5 = 10
----------
偶数列中的元素等于 (当前列 * 总列数) + 1；
例如：arra[0][2] = (2 * 5) + 1 = 11;
**重要说明：** 第一行第一列的元素为零
arr[0][0] = 1 -> 必须声明
现在，剩下的矩阵是从该列的第一个元素开始的递增或递减循环。
偶数列号的元素在每一行从该列的第一个元素开始递增1。
1 -> 第0列的第一个元素
2 -> (1 + 1)
3 -> (2 + 1).... 以此类推
4
5
----------
奇数列号的元素在每一行从该列的第一个元素开始递减1。
10 -> 第1列的第一个元素
9 -> (10 - 1)
8 -> (9 - 1).... 以此类推
7
6
英文:
I believe this is a good alternative and easy to understand. Do comment if you have a simplified version of the same.
Code:
import java.util.Arrays;
import java.util.Scanner;
public class SnakePatternProblem {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in); // *INPUT*
int row = scanner.nextInt();
int col = scanner.nextInt();
int[][] array = new int[row][col];
// Initializing first row of the array with a generalized expression
for (int i = 0; i < col; i++) {
if (i % 2 != 0) array[0][i] = col * (i+1);
else array[0][i] = (col * i) + 1;
}
array[0][0] = 1; // this element is not covered in the above loop.
// Nested loop for incrementing and decrementing as per the first row of elements.
for (int i= 1; i< row; i++){
for (int j=0; j< col; j++){
if(j%2==0) array[i][j] = array[i-1][j] + 1;
else array[i][j] = array[i-1][j] - 1;
}
}
System.out.println(Arrays.deepToString(array)); // *OUTPUT
}
}
Explanation:
Consider first row of 5 x 5 matrix named "arr" with snake pattern:
1 10 11 20 21
The element in the odd column is equivalent to ((currentColumn + 1) * Total_No_Of_columns);
Example: arr[0][1] = (1 + 1)* 5 = 10
The element in the even column is equivalent to (currentColumn * Total_No_Of_columns) + 1;
Example: arra[0][2] = (2 * 5) + 1 = 11;
Important Note: element at first row, first column is Zero
arr[0][0] = 1 -> must be declared
Now, the remaining matrix is incrementing or decrementing loop from the first element in that column.
Elements with even column number gets incremented by 1 in each row from the first element of that column.
1 -> First element of column-0
2 -> (1 + 1)
3 -> (2 + 1).... so on
4
5
Elements with odd column number gets decremented by 1 in each row from the first element of that column.
10 -> First element of column - 1
9 -> (10 - 1)
8 -> (9 - 1).... so on
7
6
答案2
得分: 0
这将生成你所描述的“蛇形”模式。
虽然可以使用三元运算符简化，但我认为这样更易读。
不过，如果有人找到更好的方法，可以在评论中分享一下。
public static int[][] genArray(int length) {
int[][] arr = new int[length][length];
int counter = 0;
for (int col = 0; col < arr.length; col++) {
if (col % 2 == 0) {
for (int row = 0; row < arr.length; row++) {
arr[row][col] = counter++;
}
} else {
for (int row = arr.length - 1; row >= 0; row--) {
System.out.println("row: " + row + ", col: " + col);
arr[row][col] = counter++;
}
}
}
return arr;
}
英文:
This will generate the "snaking" pattern you described.
It could be simplified with ternary but this makes it more readable I think
It would be interesting to find a more clever way about it though, if anyone finds a better way pls comment
public static int[][] genArray(int length) {
int[][] arr = new int[length][length];
int counter = 0;
for (int col = 0; col < arr.length; col++) {
if (col % 2 == 0) {
for (int row = 0; row < arr.length; row++) {
arr[row][col] = counter++;
}
} else {
for (int row = arr.length - 1; row >= 0; row--) {
System.out.println("row: " + row + ", col: " + col);
arr[row][col] = counter++;
}
}
}
}
return arr;
}
答案3
得分: 0
int m = 5;
int n = 4;
int[][] snakeArr = IntStream.range(0, n)
.mapToObj(i -> IntStream.range(0, m)
// even row - straight order,
// odd row - reverse order
.map(j -> (i % 2 == 0 ? j : m - j - 1) + i * m + 1)
.toArray())
.toArray(int[][]::new);
// output
Arrays.stream(snakeArr)
.map(row -> Arrays.stream(row)
.mapToObj(e -> String.format("%2d", e))
.collect(Collectors.joining(" ")))
.forEach(System.out::println);
1 2 3 4 5
10 9 8 7 6
11 12 13 14 15
20 19 18 17 16
Transposing snake array:
int[][] transposedSA = new int[m][n];
IntStream.range(0, m).forEach(i ->
IntStream.range(0, n).forEach(j ->
transposedSA[i][j] = snakeArr[j][i]));
// output
Arrays.stream(transposedSA)
.map(row -> Arrays.stream(row)
.mapToObj(e -> String.format("%2d", e))
.collect(Collectors.joining(" ")))
.forEach(System.out::println);
1 10 11 20
2 9 12 19
3 8 13 18
4 7 14 17
5 6 15 16
<details>
<summary>英文:</summary>
You can create such an array as follows:
```java
int m = 5;
int n = 4;
int[][] snakeArr = IntStream.range(0, n)
.mapToObj(i -> IntStream.range(0, m)
// even row - straight order,
// odd row - reverse order
.map(j -> (i % 2 == 0 ? j : m - j - 1) + i * m + 1)
.toArray())
.toArray(int[][]::new);
// output
Arrays.stream(snakeArr)
.map(row -> Arrays.stream(row)
.mapToObj(e -> String.format("%2d", e))
.collect(Collectors.joining(" ")))
.forEach(System.out::println);
1 2 3 4 5
10 9 8 7 6
11 12 13 14 15
20 19 18 17 16
Transposing snake array:
int[][] transposedSA = new int[m][n];
IntStream.range(0, m).forEach(i ->
IntStream.range(0, n).forEach(j ->
transposedSA[i][j] = snakeArr[j][i]));
// output
Arrays.stream(transposedSA)
.map(row -> Arrays.stream(row)
.mapToObj(e -> String.format("%2d", e))
.collect(Collectors.joining(" ")))
.forEach(System.out::println);
1 10 11 20
2 9 12 19
3 8 13 18
4 7 14 17
5 6 15 16
答案4
得分: 0
以下是翻译好的内容：
一种简单的方法是创建一个如下所示的二维数组，然后打印其转置。
1 2 3 4 5
10 9 8 7 6
11 12 13 14 15
20 19 18 17 16
21 22 23 24 25
这是一个维度为 LEN x LEN 的二维数组，其中 LEN = 5。
上述模式如下：
- 模式以
start值为1开始。 - 当主循环计数器为偶数时，将分配值给数组的计数器从
start值开始，逐次增加LEN次。最后，它将设置下一行的start以final_value_of_the_array_value_assignment_counter - 1 + LEN开始。 - 当主循环计数器为奇数时，将分配值给数组的计数器从
start值开始，逐次减少LEN次。最后，它将设置下一行的start以start + 1开始。
上述矩阵的转置如下：
1 10 11 20 21
2 9 12 19 22
3 8 13 18 23
4 7 14 17 24
5 6 15 16 25
代码方面，我们可以写成：
public class Main {
public static void main(String[] args) {
final int LEN = 5;
int[][] arr = new int[LEN][LEN];
int start = 1, j, k, col;
for (int i = 0; i < LEN; i++) {
if (i % 2 == 0) {
k = 1;
for (j = start, col = 0; k <= LEN; j++, k++, col++) {
arr[i][col] = j;
}
start = j - 1 + LEN;
} else {
k = 1;
for (j = start, col = 0; k <= LEN; j--, k++, col++) {
arr[i][col] = j;
}
start++;
}
}
// 打印矩阵的转置
for (int r = 0; r < LEN; r++) {
for (int c = 0; c < LEN; c++) {
System.out.print(arr[c][r] + "\t");
}
System.out.println();
}
}
}
输出：
1 10 11 20 21
2 9 12 19 22
3 8 13 18 23
4 7 14 17 24
5 6 15 16 25
将 LEN 的值更改为 10，您将得到以下模式：
1 20 21 40 41 60 61 80 81 100
2 19 22 39 42 59 62 79 82 99
3 18 23 38 43 58 63 78 83 98
4 17 24 37 44 57 64 77 84 97
5 16 25 36 45 56 65 76 85 96
6 15 26 35 46 55 66 75 86 95
7 14 27 34 47 54 67 74 87 94
8 13 28 33 48 53 68 73 88 93
9 12 29 32 49 52 69 72 89 92
10 11 30 31 50 51 70 71 90 91
英文:
An easy way to do it is to create a 2-D array as shown below and then print its transpose.
1 2 3 4 5
10 9 8 7 6
11 12 13 14 15
20 19 18 17 16
21 22 23 24 25
It is a 2-D array of the dimension, LEN x LEN, where LEN = 5.
The above pattern goes like this:
- The pattern starts with a
startvalue of1. - When the main loop counter is an even number, the counter, which assigns a value to the array, starts with
startvalue and increases by one,LENtimes. Finally, it setsstartfor the next row to start withfinal_value_of_the_array_value_assignment_counter - 1 + LEN. - When the main loop counter is an odd number, the counter, which assigns a value to the array, starts with
startvalue and decreases by oneLENtimes. Finally, it setsstartfor the next row to start withstart + 1.
The transpose of the above matrix will look like:
1 10 11 20 21
2 9 12 19 22
3 8 13 18 23
4 7 14 17 24
5 6 15 16 25
In terms of code, we can write it as:
public class Main {
public static void main(String[] args) {
final int LEN = 5;
int[][] arr = new int[LEN][LEN];
int start = 1, j, k, col;
for (int i = 0; i < LEN; i++) {
if (i % 2 == 0) {
k = 1;
for (j = start, col = 0; k <= LEN; j++, k++, col++) {
arr[i][col] = j;
}
start = j - 1 + LEN;
} else {
k = 1;
for (j = start, col = 0; k <= LEN; j--, k++, col++) {
arr[i][col] = j;
}
start++;
}
}
// Print the transpose of the matrix
for (int r = 0; r < LEN; r++) {
for (int c = 0; c < LEN; c++) {
System.out.print(arr[c][r] + "\t");
}
System.out.println();
}
}
}
Output:
1 10 11 20 21
2 9 12 19 22
3 8 13 18 23
4 7 14 17 24
5 6 15 16 25
Change the value of LEN to 10 and you will get the following pattern:
1 20 21 40 41 60 61 80 81 100
2 19 22 39 42 59 62 79 82 99
3 18 23 38 43 58 63 78 83 98
4 17 24 37 44 57 64 77 84 97
5 16 25 36 45 56 65 76 85 96
6 15 26 35 46 55 66 75 86 95
7 14 27 34 47 54 67 74 87 94
8 13 28 33 48 53 68 73 88 93
9 12 29 32 49 52 69 72 89 92
10 11 30 31 50 51 70 71 90 91
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