6-9 statistics of single digits (15 points)

6-9 Statistics are in single digits (15 branch )
This question requests to realize a function , It can count the number of occurrences of a bit in any integer . for example -21252 in ,2 There is 3 Time , The function should return 3.

Function interface definition ：

int Count_Digit ( const int N, const int D );

among N and D All parameters passed in by the user .N The value of is not more than int The scope of the ;

D yes [0, 9] Single digit in interval . Function must return N in D Number of occurrences .

sample input ：

-21252 2

sample output ：

3

Ideas
Pay attention to find the absolute value first , Otherwise, the remainder function cannot be calculated correctly

Sample referee test procedure ：

#include <stdio.h>

int Count_Digit ( const int N, const int D );

int main()
{
    int N, D;
	
    scanf("%d %d", &N, &D);
    printf("%d\n", Count_Digit(N, D));
    return 0;
}

/*  Your code will be embedded here  */

Test point      Tips      result      Time consuming      Memory
0    sample, Numbers <0 And appear discontinuously      The answer right     2 ms    384 KB
1     Numbers >0, Continuous occurrence      The answer right     2 ms    256 KB
2    N=0, And output 1     The answer right     2 ms    280 KB
3     Output is 0     The answer right     2 ms    364 KB
4     Statistics 0 The number of times , There may be more cycles 1     The answer right     2 ms    256 KB

Before optimization  

int Count_Digit ( const int N, const int D ){
    int tmp;
    int res;
    int arr[10]={0};
    if(N>=0){
    	res=N;
	}else{
		res=N*(-1);
	}
	if(res==D)return 1;// Handle N=0 And D=0 In special circumstances , By the way   |N|==D The situation of 
    while(res!=0){
        tmp=res%10;
        if(D==tmp){
        	arr[tmp]++;	
		}
        res=res/10;
    }
	for(int i=0;i<10;i++){
        if(arr[i]>=1){
            tmp=arr[i];
            return tmp;
        }
    }
    return 0;
}

After optimization  

int Count_Digit ( const int N, const int D ){
	int temp=N;
	if(N<0)temp*=-1;
	
	if(temp==D)return 1;// Handle N=0 And D=0 In special circumstances , By the way |N|==D The situation of 
	
	int times=0;// Number of occurrences 
	for(int i=temp;i>0;i/=10){
		if(i%10==D)times++;
	}
	return times;
}