使用数组中的前'n'个整数递归打印所有子集

Question

Recursively print the possible subsets using the first n integers of an array. Example: int[] X = [1, 2, 3, 4] and if n = 3, print [3, 2, 23, 1, 13, 12, 123]

I've sat here for hours, I've tried blindly and am now turning to you guys for some help! Here's what I have so far. Its nowhere near the answer so bear with me.

static void subsets(int[] A, int n){
    subsets("", A, n);
}
private static void subsets(String S, int[] A, int n){
    if(n == 0){
        System.out.println(S);
    } else {
        for (int i = n-1; i >= 0; i--) {
            subsets(A[n-i-1]+S, A, n-i);
        }
    }
}

Answer 1

There are many ways to solve this subsets problem, but I particularly like the one below.

It relies on the fact that, when you count from 1 to N in binary (N being a power of 2) in binary, the bits get through all the possible unique combinations. So, if you "attach" each bit to a particular value in your array, you get every possible combinations of your values.

In your example, you take N = 3 values in your array. With N bits, you can get 2^3 (or 1<<3 ) = 8 different values. So let's count from 0 to 7 in binary, and watch the bits that get on or off at each step :

• 1 = 001 -> take only the first item in your list
• 2 = 010 -> take only the second item in your list
• 3 = 011 -> ...
• 4 = 100
• 5 = 101 -> take the first and third elements in your list
• 6 = 110
• 7 = 111

And there you are : you found all possible subsets among your N elements.

Here is the code :

int[] nums = new int[]{1, 2, 3, 4};
int n = 3;

int maxValueOnNBits = 1 << n;

for (int i = 1; i <= maxValueOnNBits; i++) {
    for (int bit = 0; bit < n; bit++) {
        int mask = 1 << bit;
        if ((i & mask) == mask) System.out.print(nums[bit]);
    }
    System.out.println();
}