生成所有n位的字符串，考虑A [0..n-1]是大小为n的数组

Question

public class GenerateAllStrings {

    int []arrA;

    public GenerateAllStrings(int n)
    {
        arrA = new int[n];
    }

    public void nBits(int n)
    {
        if(n <= 0)
            System.out.println(Arrays.toString(arrA));
        else
        {
            arrA[n-1] = 0;
            nBits(n-1);
            arrA[n-1] = 1;
            nBits(n-1);
        }
    }

    public static void main(String[] args) throws java.lang.Exception
    {
        int n = 3;
        GenerateAllStrings i = new GenerateAllStrings(n);
        i.nBits(n);
    }


}

I am unable to understand recursion in this program. why is n set to 1 after printing the first set of value( I thought it should be zero)? Please explain.

Answer 1

The values you are passing are (n-1). So when it prints the first set of values, the value of n in the function is 0 but when it comes out of that recursion it is 1 since you passed the argument as (n-1) which is 0. Thus coming out of the function it gets the original value of n which was 1.

Answer 2

This program is first setting element at (n-1)th index to 0 and then calling next level recursion to print this setted value and after doing this, it again sets the (n-1)th index element to 1 and then calling next level recursion to print this setted value.

This sort of logic is happening at each level.

In order to print all the strings consists of 0 and 1 , this program is just first setting the (n-1)th value to 0( Here in the base case it's value will be printed 0) and then calling the recursion on (n-1)th value in order to print this setted value. Therefore nBits(n) is setting (n-1)th value to 0 , nBits(n-1) is setting the (n-2)th value to 0 and so on.So in the base case, 0 will be printed everywhere.

After printing the value 0000... in the base case, array element at 0th index is set to 1 in nBits(1) and then it's just printed by calling nBits(0) which prints 1000...