c - 使我的递归代码不那么复杂

Question

我正在尝试做这个问题:http://www.geeksforgeeks.org/print-all-jumping-numbers-smaller-than-or-equal-to-a-given-value/

下面的程序是我想出的，但人们理解起来并不清晰也不干净。能否修改此解决方案以使其不那么复杂？

#include <stdio.h>

int calculate_digits(int x)
{
    if (x <= 9)
        return 1;
    int y = calculate_digits(x/10);
    return y+1;
}

int add_up(int *a, int size, int can_add)
{
    int i, sum=0, mul = 1;

    for (i=0;i<=size;i++) {
        sum = sum*mul + a[i];
        mul = 10;
    }
    return sum*10+can_add;
}

int j;
int output[100];
int foo(int x, int index , int value, int size, int j)
{
    if (x <= 10 || value >= 10)
        return 11;
    if (index >= size) {
        int i;
        for (i=0;i<j;i++)
            printf("%d", output[i]);
        printf("\n");
        printf("next output\n\n");
        return 0;
    }

    if (index == 0) {
        int k;
        for (k=1;k<=9;k++) {
            output[0] = k;
            foo(x, 1, k+1, size, 1);
            foo(x, 1, k-1, size, 1);
        }
    } else if (abs(output[index-1]-value) == 1){
            if (add_up(output, index-1, value) <= x) {
                output[index] = value;
                foo(x, index+1, value+1, size, j+1);
            } else {
                foo(x, index+1, value+1, size, j);
            }
    } else
        foo(x, index+1, value, size, j);
}

int main(void) {
    printf("%d\n", foo(105, 0, 0, calculate_digits(105), 0));
    return 0;
}

我正在寻找任何更好的递归解决方案，或者如果可能，有人可以修改此代码并使其变得更好。

Answer 1

我想你正在寻找类似的东西:

void jumpingc(int current, int max) {
    if (current <= max) {
        printf("%d ", current);
        int ld = current % 10;
        if (ld > 0)
            jumpingc(current * 10 + ld - 1, max);
        if (ld < 9)
            jumpingc(current * 10 + ld + 1, max);
    }
}

void jumping(int max) {
    printf("%d ", 0);
    for (int i = 1; i <= 9; i++)
        jumpingc(i, max);
}

int main(void) {
    int max = 105;
    jumping(max);
    return 0;
}