有人能给我解释一下这段代码吗？关于用户输入、扫描仪等

Question

有人能给我解释一下这个代码是怎么工作的吗

它允许用户输入数字，直到1000，然后打印原始输入的数字，偶数和奇数，全部在一个单独的数组中。但是我不明白当它输出偶数和奇数而不是偶数和奇数的数目时，有gem++和gem1++的部分

在把这个

double even[] = new double[gem];
double odd[] = new double [gem1]; 

为什么它需要再次重复gem=0和gem1=0？如果我问了太多的问题，我很抱歉，我只是感到困惑，我上周刚学了java

public class wutt {
    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);
        System.out.print("Enter no. of elements you want in array : ");
        int n = s.nextInt();

        if (1 <= n && n <= 1000) {

            double a[] = new double[n];
            int gem = 0, gem1 = 0;

            System.out.println("Enter all the elements : ");
            for (int i = 0; i < n; i++) {
                a[i] = s.nextInt();
                if (a[i] % 2 == 0)
                    gem++;
                else
                    gem1++;
            }

            double even[] = new double[gem];
            double odd[] = new double[gem1];

            gem = 0;
            gem1 = 0;

            for (int i = 0; i < a.length; i++) {
                if (a[i] % 2 == 0) {
                    even[gem] = a[i];
                    gem++;
                } else {
                    odd[gem1] = a[i];
                    gem1++;
                }
            }

            System.out.println("Original: " + Arrays.toString(a));
            System.out.println("Odd: " + Arrays.toString(odd));
            System.out.println("Even: " + Arrays.toString(even));

        } else
            System.out.print("Invalid input");
    }
}

Answer 1

如果希望在用户输入大于1000或小于0的数字后程序停止，则需要在if条件中添加break语句

if (size < 0 || size > 1000) {
    System.out.println("Size must be between 0 and 1000");
    break;
}

Answer 2

在double even[] = new double[gem]; double odd[] = new double [gem1];之前的代码试图获取发生的几率数和偶数数，并将所有输入的元素放入数组a。 在所有这些之后，现在我们得到的是一个名为a的数字数组，包含所有输入的元素。两个数字分别是gem和gem1，其中包含发生的几率和偶数。 所以 我们得到gem（numberOfEvens）、gem1（numberoftoffics）和list a

接下来，我们需要将a的所有赔率放入一个名为odd[]的新数组，大小为gem1，然后 将a中的所有even放入一个名为偶数[]且大小为gem的新数组。此时，变量gem1和gem的任务完成了。它们变得毫无用处

现在，我们需要遍历列表，选择奇数和偶数，然后按顺序将它们逐个放入数组中。这就是为什么我们需要两个初始化为0的新变量。 在这种情况下，因为gem和gem1是无用的，所以它们被重新分配来帮助操作树数组a、odd[]和even[]

Answer 3

因此，用户在数组中输入他/她想要的元素数（n）

double a[] = new double[n];   // a is the array that is initialised to accommodate n elements
int gem = 0, gem1 = 0;        // gem is the counter for "even" numbers and "gem1" the counter for odd numbers, and like every good counter, they start at 0

System.out.println("Enter all the elements : ");
for (int i = 0; i < n; i++) {     // so we ask the user to input n elements
    a[i] = s.nextInt();           // here we read every input and put it in the a array
    if (a[i] % 2 == 0)            // if the new number is even
        gem++;                    // we increase the even counter "gem"
    else                          // otherwise, when it is an odd number
        gem1++;                   // we increase the odd counter
}

double even[] = new double[gem];  // now we create a new array where we want to hold all the even numbers, we do that by telling it how many even numbers we have counted before (gem)
double odd[] = new double[gem1];  // and a new array for all odd numbers (gem1 was our counter)

gem = 0;                          // now we reinitialise the counters, because we want to start from the beginning
gem1 = 0;

for (int i = 0; i < a.length; i++) { // in order to copy all numbers from the a array into the two other arrays for even and odd numbers, we iterate over the whole length of the a array. i is the index for the "a" array
    if (a[i] % 2 == 0) {             // ever even number we encounter
        even[gem] = a[i];            // we put in the even array
        gem++;                       // while gem, the "even numbers counter" is our index for the "even" array
    } else {
        odd[gem1] = a[i];            // odd numbers are for the odd array
        gem1++;                      // while the former "odd numbers counter" now serves as our "odd" array index
    }
}

差不多就是这样。首先，用户在一个数组中输入所有数字，然后简单地计算输入的奇数和偶数

然后创建两个新数组，一个用于偶数，另一个用于奇数，由于我们计算了它们，我们知道这两个新数组必须有多大

最后，所有的数字再次被迭代，并放入相应的数组中。 最后有3个数组，一个包含所有数字，一个包含偶数数字，另一个只包含奇数数字

编辑

以下是在不改变该方法性质的情况下可以进行的一些小改动：

double allNumbers[] = new double[n];  // "allNumbers" is way more specific than "a"
int oddCounter = 0;                   // "oddCounter" instead of "gem"
int evenCounter = 0;                  // numbers in variables like "gem1" is really bad practice, because numbers don't say anything about the nature of the variable

System.out.println("Enter all the elements : ");

for (int i = 0; i < n; i++) {
    allNumbers[i] = s.nextInt();
    if (allNumbers[i] % 2 == 0) {
        evenCounter++;
    } else {
        oddCounter++;
    }
}
// until here nothing changes but the names

double even[] = new double[evenCounter];
double odd[] = new double[oddCounter];

int oddIndex = 0;                   // and here we create new variables, instead of reusing old ones
int evenIndex = 0;                  // there is absolutely no performance gain in reusing primitives like this - it's just confusing
for (int i = 0; i < allNumbers.length; i++) {
    if (allNumbers[i] % 2 == 0) {
        even[evenIndex++] = allNumbers[i];   // the "++" can be done directly in the first expression. that's just to make it shorter.
    } else {
        odd[oddIndex++] = allNumbers[i];     // it is not more performant nor easier to read - just shorter
    }
}

编辑（再次）

这就是数组的样子，比如当你输入4个数字时：

gem = 0
gem1 = 0
n = 4               // user said 4 
a = [ , , , ]       // array a is empty but holds the space for 4 numbers

a = [1, , , ]       // user enters 1
     ^
    i=0

gem1 = 1            // 1 is an odd number -> gem1++

a = [1,4, , ]       // user entered "4"
       ^
      i=1
gem = 1             // 4 is an even number -> gem++

a = [1,4,2, ]       // user entered "2"
         ^
        i=2
gem = 2             // 24 is an even number -> gem++

a = [1,4,2,7]       // user entered "7"
           ^
          i=3
gem1 = 2             // 7 is an odd number -> gem1++


then we fill the other arrays

even = [ , ]          // gem is 2, so we have 2 even numbers
odd  = [ , ]          // gem1 is 2, so we have 2 odd numbers

a = [1,4,2,7]
     ^ 
    i=0

odd[1, ]    // for i=0, a[i] is 1, which is an odd number

a = [1,4,2,7]
       ^ 
      i=1

even = [4, ]    // for i=1, a[i] is 4, which is an even number

a = [1,4,2,7]
         ^ 
        i=2

even = [4,2]    // for i=2, a[i] is 2, which is an even number

a = [1,4,2,7]
           ^ 
          i=3

odd = [1,7]    // for i=3, a[i] is 7, which is an odd number

and in the end you have

a    = [1,4,2,7]
even = [4,2]
odd  = [1,7]