C中的递归问题（数组中奇数的总和）

Question

I have questions about a recursion function. The program is supposed to compute the sum of n odd integers and the point is each time we find the sum, we must print the following:

For example, If the user gives 5 numbers (5,4,3,2,1), the list will be:

5 4 3 2 1

Then it will print:

5[0] (Yes) 4[5] (Yes) 3[5] (Yes) 2[8] (No) 1[8] (No)

At first, it prints the number, after that in the [] the sum of the next odd numbers and at the end in parenthesis, a (Yes) or (No) if the number^2 in each node is greater from the sum of its next odd numbers. Like in the first one, 5>0 so it's a Yes.

I've written this so far, I can't print it though, is there anything missing?

int checkSumOfOdds(struct list *ptr) {
    int k=0;

    if (ptr != NULL){
        k = checkSumOfOdds(p->next);
        if((ptr->next)^2 > k){
            printf(" %d [%d] (YES)",ptr->value, k);}
        else{
            printf(" %d [%d] (YES)",ptr->value, k);}

        k += ptr->value;
        return k;
    }

    return 0;
}

Answer 1

According to the problem statement, the values to be displayed are from the first input to the last. '5' comes first, then '4'...

This determines how the recursive function behaves. Does it call itself first, or does it at the end? In your case the head values are to be dealt with first, thus the recursive call is to be done after the number is processed. At the end.

Then your test if (ptr != NULL) should only affect the recursive call, not the whole body.

The ^ operator is exclusive OR in C... Probably not what you want (though it is possible to use it, but not like that). To test if a number is odd, simply check its first bit (with & 1 ), or check if N % 2 is > 0 .

Since the number is processed and displayed early (because the recursive call is at the end), you can carry the sum with the calls.

I'm guessing the sum to be displayed should start with the current number (or, as an exercise, simply delay the summing...).

An example of code that should work

int checkSumOfOdds(struct list *ptr, int sum) {
     int isodd = ptr->value & 1;
     if (isodd) sum += ptr->value;
     printf("%d[%d](%s) ", ptr->value, sum, isodd ? "Yes":"No");
     return ptr->next ? checkSumOfOdds(ptr->next, sum) : sum;
}

to be called

 checkSumOfOdds(headOfList, 0);
 printf("\n");