Таким образом, пользователь вводит количество элементов, которое он / она хочет в массиве (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
}
}
, и это почти все. Сначала пользователь вводит все числа в одном массиве и просто подсчитывает, сколько нечетных и сколько четных чисел было введено,
, затем создаются два новых массива, один для четного и один для нечетных чисел, и так как мыпосчитав их, мы знаем, насколько большими должны быть эти два новых массива.
И, наконец, все числа снова повторяются и помещаются в соответствующий им массив. В конце у вас есть 3 массива, один из которых содержит все чисел, один из которых содержит четные числа и один содержит только нечетные числа.
РЕДАКТИРОВАТЬ
Вот несколько незначительных изменений, которые вы можете сделать, не меняя природу этого метода:
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
}
}
РЕДАКТИРОВАТЬ (снова)
Вот так выглядят массивы, скажем, когда вы вводите 4 цифры:
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]