x=3 y=3
用户输入：
101
100
000

import java.util.Scanner;
public class RedVsGreen {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int x, y;
        String line;
        String[] lineVector = new String[3];
        while (lineVector.length != 2 || (Integer.parseInt(lineVector[0]) >= 1000 || Integer.parseInt(lineVector[0]) <= 1)
                || (Integer.parseInt(lineVector[1]) <= 1 || Integer.parseInt(lineVector[1]) >= 1000)) {
            System.out.print("请输入x和y，用逗号分隔（大于1且小于1000）：");
            // 读取x和y
            line = scanner.nextLine();
            // 通过逗号分隔所有值
            lineVector = line.split("\\s*,\\s*");
        }
        // 解析值为整数
        x = Integer.parseInt(lineVector[0]);
        y = Integer.parseInt(lineVector[1]);
        for (int i = 0; i < x; i++) {
        }
    }
}

**接收 x 和 y**
在这个问题中，你的 while 条件过于复杂。使用 do while 格式更适合，因为你至少会获取一次值，而且代码更可读。
另外，你可以使用 x 来代替 `Integer.parseInt(lineVector[0])`（y 也是同样的道理），而不是重复这个过程，这样可以缩短条件判断。
```java
String line;
String[] lineVector;
int x = -1, y = -1;
do {
    System.out.print("请输入 x 和 y，以逗号分隔（大于1且小于1000）：");
    line = scanner.nextLine();
    lineVector = line.split("\\s*,\\s*");
    if (lineVector.length != 2)
        continue;
    x = Integer.parseInt(lineVector[0]);
    y = Integer.parseInt(lineVector[1]);
} while (!((x > 1 && x < 1000 && y > 1 && y < 1000)));

public static boolean isBinary(String s) {
    for (int i = 0; i < s.length(); ++i) {
        char c = s.charAt(i);
        if (c != '0' && c != '1')
            return false;
    }
    return true;
}

String[] binaries = new String[x];
System.out.println("输入长度为 " + y + " 的二进制值：");
// 获取二进制值
for (int i = 0; i < x; ++i) {
    binaries[i] = scanner.nextLine();
    while (!isBinary(binaries[i]) || binaries[i].length() != y) {
        System.out.println("无效的二进制值。请重新输入新值：");
        binaries[i] = scanner.nextLine();
    }
}

int[][] field = new int[x][y];
for (int i = 0; i < x; ++i) {
    System.out.println();
    for (int j = 0; j < y; ++j) {
        char c = binaries[i].charAt(j);
        if (c == '0')
            field[i][j] = 0;
        else
            field[i][j] = 1;
        System.out.print(field[i][j] + " ");
    }
}

<details>
<summary>英文:</summary>
**Receiving x and y**
    while ( lineVector.length !=2 ...)
Your while condition is too complicated, a do while format matches this problem much better since you are going to get the values at least once, and the code is more readable.
Also you could have used x instead of `Integer.parseInt(lineVector[0])` (same for y) instead of repeating the process, this would have shortened the condition.
    
    String line;
    String[] lineVector;
    int x = -1, y = -1;
        do {
            System.out.print(&quot;Please enter x and y, comma separated (more than 1 and less than 1000):&quot;);
            line = scanner.nextLine();
            lineVector = line.split(&quot;\\s*,\\s*&quot;);
            if(lineVector.length != 2)
                continue;
            x = Integer.parseInt(lineVector[0]);
            y = Integer.parseInt(lineVector[1]);
        } while (!((x &gt; 1 &amp;&amp; x &lt; 1000 &amp;&amp; y &gt; 1 &amp;&amp; y &lt; 1000)));
First of all I removed the initial value of `lineVector` as it is unnecessary to initialize it in this case (you needed to do it because of it being present in your while condition).
I initialized `x` and `y` to -1 (Any number not in our range would work) in order to make sure the do-while condition is fulfilled until proper values are offered for both of the numbers.
**Function to check binary values**
Create a function to check if string values are binary.
    public static boolean isBinary(String s) {
        for(int i = 0; i &lt; s.length(); ++i) {
            char c = s.charAt(i);
            if(c != &#39;0&#39; &amp;&amp; c != &#39;1&#39;)
                return false;
        }
        return true;
    }
**Receive binaries**
    String[] binaries = new String[x];
    
    System.out.println(&quot;Enter binaries with length &quot; + y + &quot; :&quot;);
    
    //Get binaries
    for(int i = 0; i &lt; x; ++i) {
        binaries[i] = scanner.nextLine();
    
        while(!isBinary(binaries[i]) || binaries[i].length() != y) {
            System.out.println(&quot;Invalid binary value. Re-enter new value:&quot;);
            binaries[i] = scanner.nextLine();
        }
    }
Receive binary values and continue asking if invalid.
**Filling the 2D array**
    int[][] field = new int[x][y];
    for(int i = 0; i &lt; x; ++i) {
        System.out.println();
        for(int j = 0; j &lt; y; ++j) {
            char c = binaries[i].charAt(j);
            if(c == &#39;0&#39;)
                field[i][j] = 0;
            else
                field[i][j] = 1;
            System.out.print(field[i][j] + &quot; &quot;);
        }
    }
`Integer.parseInt` is not necessary here as there are only two possible values (0 and 1)
 
</details>