如果 integer 在数组中仅存在一次，则返回 true 的递归方法

Question

For example:

int[] arr = {1,2,2,3,4,5};

3 exists only once the method should return true , 2 exists twice the method should return false , 0 doesn't exist the method should return false . I keep getting a runtime error, how do I fix it or is there any way to the method (recursions only). Here is my code:

public static boolean list(int[] a, int num) {
    return helper(a, 0, num);
}

public static boolean helper(int[] a, int i, int num) {
    int count = 0;
    if (a[i] == num)
        count++;
    if (count == 1 && i == a.length)
        return true;
    else if (count != 1 && i == a.length)
        return false;
    else
        return helper(a, i++, num);
}

Answer 1

I ran your code and got a java.lang.StackOverflowError . That indicates that your recursive method, named helper keeps calling itself infinitely. It does that because neither of the conditions in the if statements that precede the return statements are ever true.

You don't want to recursively call the method when one of the following two conditions are true:

1. You have encountered a second occurrence of the number you are searching for in the array.
2. You have reached the end of the array.

If you keep initializing count to zero in every call to method helper , it will never be greater than one. Hence you should make it a parameter of method helper .

You just need to check whether i equals a.length to determine whether you have reached the end of the array.

/**
 * Determines whether 'num' occurs exactly once in 'a'.
 *
 * @param a     - array to search
 * @param count - number of occurrences of 'num' in 'a'
 * @param i     - index in 'a'
 * @param num   - number to search for
 *
 * @return 'true' if 'num' occurs exactly once in 'a', otherwise 'false'.
 */
public static boolean helper2(int[] a, int count, int i, int num) {
    if (count > 1) {
        return false;
    }
    if (i == a.length) {
        return count == 1;
    }
    if (a[i] == num) {
        count++;
    }
    return helper2(a, count, i + 1, num);
}

And the initial call to the method is (for example, if you are checking whether the number 3 appears only once in the array.

int[] a = new int[]{1, 2, 2, 3, 4, 5}; // array of integers to search
int count = 0; // number of occurrences
int index = 0; // index in 'a'
int num = 3; // number to search for
boolean single = helper2(a, count, index, num);

Answer 2

This is much easier to do with an iterative method :

public static boolean iterativeMethod(int[] arr, int number) {
    // filter the desired numbers in the array
    // and check if their quantity is equal to '1'
    return Arrays.stream(arr).filter(i -> i == number).count() == 1;
}

public static void main(String[] args) {
    int[] arr = {1, 2, 2, 3, 4, 5};
    System.out.println(iterativeMethod(arr, 3)); // true
    System.out.println(iterativeMethod(arr, 2)); // false
    System.out.println(iterativeMethod(arr, 0)); // false
}