打印非连续方式的数量

Question

Given a positive integer N, print all integers between 1 and 2^N such that there is no consecutive 1's in its Binary representation.

I have below code but it is printing duplicate sometimes. Is it possible to print without duplicates?

#include <stdio.h>

int a[100];
void foo(int i, int size)
{
    if (i >= size) {
        int i;
        for (i=0;i<size;i++)
            printf("%d\n", a[i]);
        printf("----\n");
        return;
    }

    if (a[i-1] == 1 || a[i-1] == 0)
        a[i] = 0;
    foo(i+1, size);
    if (a[i-1] == 0)
        a[i] = 1;   
    foo(i+1, size);
}

int main(void) {
    int i = 0;
    int size = 5;
    a[i] = 1;
    foo(1, size);
    return 0;
}

I have this http://ideone.com/cT4Hco python program which uses hash maps to print the elements but I think we can do this without hashmaps also.

Answer 1

Couple of notes:

• you shouldn't start the backtracking from index 1. Instead, start from 0 since your numbers would be in the range [0, n-1] in array a
• you shouldn't initialize a[0] to 1 since a[0] = 0 is also a valid case.
• if (a[i-1] == 1 || a[i-1] == 0) is redundant

Code:

#include <stdio.h>

int a[100];
void foo(int i, int size)
{
    if (i >= size) {
        int i;
        for (i=0;i<size;i++)
            printf("%d ", a[i]);
        printf("\n----\n");
        return;
    }

    a[i] = 0;
    foo(i+1, size);
    if ( i == 0 || a[i-1] == 0) {
        a[i] = 1;
        foo(i+1, size);
    }
}

int main(void) {
    int i = 0;
    int size = 5;
    foo(0, size);
    return 0;
}

You might also want to filter the solution 0 0 0 ... 0 during the printing since you need only the numbers from 1 to 2^n . If 2^n is included you should also print it. The backtracking considers the numbers 0, ...., 2^n-1